A Mathematical Olympiad ApproachRadmila Bulajich Manfrino [604432]

Licență

A Mathematical Olympiad ApproachRadmila Bulajich Manfrino [604432]

Byadmin ianuarie 1, 2024

Inequalities
A Mathematical Olympiad ApproachRadmila Bulajich Manfrino
José Antonio Gómez OrtegaRogelio Valdez Delgado
Birkhäuser
Basel · Boston · Berlin

2000 Mathematical Subject Classiﬁcation 00A07; 26Dxx, 51M16
Library of Congress Control Number: 2009929571Bibliograﬁsche Information der Deutschen Bibliothek
Die Deutsche Bibliothek verzeichnet diese Publikation in der Deutschen National-bibliograﬁe; detaillierte bibliograﬁsche Daten sind im Internet über <http://dnb.ddb.de> abrufbar.
ISBN 978-3-0346-0049-1 Birkhäuser Verlag, Basel – Boston – Berlin
This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, re-use of illustra-tions, recitation, broadcasting, reproduction on microﬁlms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained.
© 2009 Birkhäuser Verlag AG
Basel · Boston · BerlinPostfach 133, CH-4010 Basel, SchweizEin Unternehmen von Springer Science+Business MediaGedruckt auf säurefreiem Papier, hergestellt aus chlorfrei gebleichtem Zellstoff. TCF ∞Printed in Germany
ISBN 978-3-0346-0049-1 e-ISBN 978-3-0346-0050-79 8 7 6 5 4 3 2 1 www.birkhauser.chAutors:
Radmila Bulajich Manfrino
Rogelio Valdez DelgadoFacultad de Ciencias Universidad Autónoma Estado de MorelosAv. Universidad 1001Col. Chamilpa62209 Cuernavaca, MorelosMéxicoe-mail: [anonimizat] valdez@uaem.mxJosé Antonio Gómez OrtegaDepartamento de MatemàticasFacultad de Ciencias, UNAMUniversidad Nacional Autónoma de MéxicoCiudad Universitaria04510 México, D.F .Méxicoe-mail: [anonimizat]

Introduction
This book is intended for the Mathematical Olympiad students who wish to pre-
pare for the study of inequalities, a topic now of frequent use at various levelsof mathematical competitions. In this volume we present both classic inequalities
and the more useful inequalities for confronting and solving optimization prob-
lems. An important part of this book deals with geometric inequalities and thisfact makes a big diﬀerence with respect to most of the books that deal with this
topic in the mathematical olympiad.
The book has been organized in four chapters which have each of them a
diﬀerent character. Chapter 1 is dedicated to present basic inequalities. Most of
them are numerical inequalities generally lacking any geometric meaning. How-ever, where it is possible to provide a geometric interpretation, we include it as
we go along. We emphasize the importance of some of these inequalities, such as
the inequality between the arithmetic mean and the geometric mean, the Cauchy-
Schwarz inequality, the rearrangement inequality, the Jensen inequality, the Muir-
head theorem, among others. For all these, besides giving the proof, we presentseveral examples that show how to use them in mathematical olympiad prob-
lems. We also emphasize how the substitution strategy is used to deduce several
inequalities.
The main topic in Chapter 2 is the use of geometric inequalities. There we ap-
ply basic numerical inequalities, as described in Chapter 1, to geometric problemsto provide examples of how they are used. We also work out inequalities which
have a strong geometric content, starting with basic facts, such as the triangle
inequality and the Euler inequality. We introduce examples where the symmetri-cal properties of the variables help to solve some problems. Among these, we pay
special attention to the Ravi transformation and the correspondence between an
inequality in terms of the side lengths of a triangle a,b,cand the inequalities
that correspond to the terms s,randR, the semiperimeter, the inradius and the
circumradius of a triangle, respectively. We also include several classic geometricproblems, indicating the methods used to solve them.
In Chapter 3 we present one hundred and twenty inequality problems that
have appeared in recent events, covering all levels, from the national and up tothe regional and international olympiad competitions.

vi Introduction
In Chapter 4 we provide solutions to each of the two hundred and ten exer-
cises in Chapters 1 and 2, and to the problems presented in Chapter 3. Most of
the solutions to exercises or problems that have appeared in international math-ematical competitions were taken from the oﬃcial solutions provided at the time
of the competitions. This is why we do not give individual credits for them.
Some of the exercises and problems concerning inequalities can be solved us-
ing diﬀerent techniques, therefore you will ﬁnd some exercises repeated in diﬀerent
sections. This indicates that the technique outlined in the corresponding sectioncan be used as a tool for solving the particular exercise.
The material presented in this book has been accumulated over the last ﬁf-
teen years mainly during work sessions with the students that won the nationalcontest of the Mexican Mathematical Olympiad. These students were develop-
ing their skills and mathematical knowledge in preparation for the internationalcompetitions in which Mexico participates.
We would like to thank Rafael Mart´ ınez Enr´ ıquez, Leonardo Ignacio Mart´ ınez
Sandoval, David Mireles Morales, Jes´ us Rodr´ ıguez Viorato and Pablo Sober´ on
Bravo for their careful revision of the text and helpful comments for the improve-
ment of the writing and the mathematical content.

Contents
Introduction vii
1 Numerical Inequalities 1
1 . 1 O r d e r i n t h e r e a l n u m b e r s………………….. 1
1.2 The quadratic function ax2+2bx+c…………….. 4
1.3 A fundamental inequality,
a r i t h m e t i c m e a n – g e o m e t r i c m e a n ………………. 7
1.4 A wonderful inequality:
T h e r e a r r a n g e m e n t i n e q u a l i t y………………… 1 3
1 . 5 C o n v e x f u n c t i o n s………………………. 2 01 . 6 A h e l p f u l i n e q u a l i t y …………………….. 3 31 . 7 T h e s u b s t i t u t i o n s t r a t e g y ………………….. 3 91 . 8 M u i r h e a d ’ s t h e o r e m …………………….. 4 3
2 Geometric Inequalities 51
2 . 1 T w o b a s i c i n e q u a l i t i e s ……………………. 5 12 . 2 I n e q u a l i t i e s b e t w e e n t h e s i d e s o f a t r i a n g l e………….. 5 42.3 The use of inequalities in the geometry of the triangle . . . . . . . 592 . 4 E u l e r ’ s i n e q u a l i t y a n d s o m e a p p l i c a t i o n s…………… 6 62.5 Symmetric functions of a,bandc………………. 7 0
2.6 Inequalities with areas and perimeters . . . . . . . . . . . . . . . . 752.7 Erd˝ o s – M o r d e l l T h e o r e m …………………… 8 0
2 . 8 O p t i m i z a t i o n p r o b l e m s……………………. 8 8
3 Recent Inequality Problems 101
4 Solutions to Exercises and Problems 117
4 . 1 S o l u t i o n s t ot h e e x e r c i s e s i n C h a p t e r 1……………. 1 1 74 . 2 S o l u t i o n s t ot h e e x e r c i s e s i n C h a p t e r 2……………. 1 4 04 . 3 S o l u t i o n s t ot h e p r o b l e m s i n C h a p t e r 3 ……………. 1 6 2
Notation 205

viii Contents
Bibliography 207
Index 209

Chapter 1
Numerical Inequalities
1.1 Order in the real numbers
A very important property of the real numbers is that they have an order. The
order of the real numbers enables us to compare two numbers and to decide which
one of them is greater or whether they are equal. Let us assume that the realnumbers system contains a set P, which we will call the set of positive numbers,
and we will express in symbols x>0i fxbelongs to P. We will also assume the
following three properties.
Property 1.1.1. Every real number xhas one and only one of the following prop-
erties:
(i)x=0,
(ii)x∈P(that is, x>0),
(iii)−x∈P(that is, −x>0).
Property 1.1.2. Ifx,y∈P,t h e n x+y∈P(in symbols x>0,y>0⇒x+y>0).
Property 1.1.3. Ifx,y∈P,t h e n xy∈P(in symbols x>0,y>0⇒xy >0).
If we take the “real line” as the geometric representation of the real numbers,
by this we mean a directed line where the number “0”has been located and serves
to divide the real line into two parts, the positive numbers being on the side
containing the number one “1”. In general the number one is set on the right hand
side of 0. The number 1 is positive, because if it were negative, since it has theproperty that 1 ·x=xfor every x, we would have that any number x/negationslash=0w o u l d
satisfy x∈Pand−x∈P, which contradicts property 1.1.1.
Now we can deﬁne the relation ais greater than bifa−b∈P(in symbols
a>b). Similarly, ais smaller than bifb−a∈P(in symbols a<b). Observe that

2 Numerical Inequalities
aa. We can also deﬁne that ais smaller than or equal to
bifa0,t h e n ac < bc .
In fact, to prove (i) we see that a+c0⇔
b−a>0⇔a<b. To prove (ii), we proceed as follows: a0a n d
sincec>0, then ( b−a)c>0, therefore bc−ac >0a n dt h e n ac < bc .
Exercise 1.1. Given two numbers aandb, exactly one of the following assertions
is satisﬁed, a=b,a>b ora<b.
Exercise 1.2. Prove the following assertions.
(i)a<0,b<0⇒ab >0.
(ii)a<0,b>0⇒ab <0.
(iii)a<b,b<c⇒a<c.
(iv)a<b,c<d⇒a+c<b+d.
(v)a<b⇒− b<−a.
(vi)a>0⇒1
a>0.
(vii)a<0⇒1
a<0.
(viii) a>0,b>0⇒a
b>0.
(ix) 0 <a1⇒a2>a .
(xi) 0 <a< 1⇒a2<a .
Exercise 1.3. (i) If a>0,b>0a n d a2<b2,t h e n a<b.
(ii) If b>0, we have thata
b>1 if and only if a>b.
The absolute value of a real number x, which is denoted by |x|, is deﬁned as
|x|=/braceleftBigg
xifx≥0,
−xifx<0.
Geometrically, |x|is the distance of the number x(on the real line) from the origin
0. Also, |a−b|is the distance between the real numbers aandbon the real line.

1.1 Order in the real numbers 3
Exercise 1.4. For any real numbers x,aandb, the following hold.
(i)|x|≥0, and is equal to zero only when x=0 .
(ii)|−x|=|x|.
(iii)|x|2=x2.
(iv)|ab|=|a||b|.
(v)/vextendsingle/vextendsingle/vextendsinglea
b/vextendsingle/vextendsingle/vextendsingle=|a|
|b|,w i t h b/negationslash=0 .
Proposition 1.1.5 (Triangle inequality). The triangle inequality states that for any
pair of real numbers aandb,
|a+b|≤|a|+|b|.
Moreover, the equality holds if and only if ab≥0.
Proof. Both sides of the inequality are positive; then using Exercise 1.3 it is suﬃ-
cient to verify that |a+b|2≤(|a|+|b|)2:
|a+b|2=(a+b)2=a2+2ab+b2=|a|2+2ab+|b|2≤|a|2+2|ab|+|b|2
=|a|2+2|a||b|+|b|2=(|a|+|b|)2.
In the previous relations we observe only one inequality, which is obvious since
ab≤|ab|.N o t et h a t ,w h e n ab≥0, we can deduce that ab=|ab|=|a||b|,a n dt h e n
the equality holds. /square
Thegeneral form of the triangle inequality for real numbers x1,x2,…,x n,
is
|x1+x2+···+xn|≤|x1|+|x2|+···+|xn|.
The equality holds when all xi’s have the same sign. This can be proved in a similar
way or by the use of induction. Another version of the last inequality, which isused very often, is the following:
|±x
1±x2±···± xn|≤|x1|+|x2|+···+|xn|.
Exercise 1.5. Letx,y,a,bbe real numbers, prove that
(i)|x|≤b⇔−b≤x≤b,
(ii)||a|−|b|| ≤ |a−b|,
(iii)x2+xy+y2≥0,
(iv)x>0,y>0⇒x2−xy+y2>0.
Exercise 1.6. For real numbers a,b,c,p r o v et h a t
|a|+|b|+|c|−|a+b|−|b+c|−|c+a|+|a+b+c|≥0.

4 Numerical Inequalities
Exercise 1.7. Leta,bbe real numbers such that 0 ≤a≤b≤1. Prove that
(i) 0≤b−a
1−ab≤1,
(ii) 0≤a
1+b+b
1+a≤1,
(iii) 0 ≤ab2−ba2≤1
4.
Exercise 1.8. Prove that if n,mare positive integers, thenm
n<√
2 if and only if√
2<m+2n
m+n.
Exercise 1.9. Ifa≥b,x≥y,t h e n ax+by≥ay+bx.
Exercise 1.10. Ifx,y>0, then/radicalBig
x2
y+/radicalBig
y2
x≥√
x+√
y.
Exercise 1.11. (Czech and Slovak Republics, 2004) Let a,b,c,dbe real numbers
witha+d=b+c,p r o v et h a t
(a−b)(c−d)+(a−c)(b−d)+(d−a)(b−c)≥0.
Exercise 1.12. Letf(a,b,c,d )=(a−b)2+(b−c)2+(c−d)2+(d−a)2.F o r
a<b<c<d ,p r o v et h a t
f(a,c,b,d )>f(a,b,c,d )>f(a, b, d, c ).
Exercise 1.13. (IMO, 1960) For which real values of xthe following inequality
holds:
4×2
(1−√
1+2x)2<2x+9 ?
Exercise 1.14. Prove that for any positive integer n, the fractional part of√
4n2+n
is smaller than1
4.
Exercise 1.15. (Short list IMO, 1996) Let a,b,cbe positive real numbers such
thatabc=1 .P r o v et h a t
ab
a5+b5+ab+bc
b5+c5+bc+ca
c5+a5+ca≤1.
1.2 The quadratic function ax2+2bx+c
One very useful inequality for the real numbers is x2≥0, which is valid for any
real number x(it is suﬃcient to consider properties 1.1.1, 1.1.3 and Exercise 1.2
of the previous section). The use of this inequality leads to deducing many other
inequalities. In particular, we can use it to ﬁnd the maximum or minimum of a
quadratic function ax2+2bx+c. These quadratic functions appear frequently in
optimization problems or in inequalities.

1.2 The quadratic function ax2+2bx+c 5
One common example consists in proving that if a>0, the quadratic function
ax2+2bx+cwill have its minimum at x=−b
aand the minimum value is c−b2
a.
In fact,
ax2+2bx+c=a/parenleftbigg
x2+2b
ax+b2
a2/parenrightbigg
+c−b2
a
=a/parenleftbigg
x+b
a/parenrightbigg2
+c−b2
a.
Since/parenleftbig
x+b
a/parenrightbig2≥0 and the minimum value of this expression, zero, is attained
when x=−b
a, we conclude that the minimum value of the quadratic function is
c−b2
a.
Ifa<0, the quadratic function ax2+2bx+cwill have a maximum at x=−b
a
and its value at this point is c−b2
a.I nf a c t ,s i n c e ax2+2bx+c=a/parenleftbig
x+b
a/parenrightbig2+c−b2
a
and since a/parenleftbig
x+b
a/parenrightbig2≤0 (because a<0), the greatest value of this last expression
is zero, thus the quadratic function is always less than or equal to c−b2
aand
assumes this value at the point x=−b
a.
Example 1.2.1. Ifx,yare positive numbers with x+y=2a, then the product xy
is maximal when x=y=a.
Ifx+y=2a,t h e n y=2a−x. Hence, xy=x(2a−x)=−x2+2ax=
−(x−a)2+a2has a maximum value when x=a,a n dt h e n y=x=a.
This can be interpreted geometrically as “of all the rectangles with ﬁxed
perimeter, the one with the greatest area is the square” .I nf a c t ,i f x,yare the
lengths of the sides of the rectangle, the perimeter is 2( x+y)=4a,a n di t sa r e a
isxy, which is maximized when x=y=a.
Example 1.2.2. Ifx,yare positive numbers with xy=1,t h es u m x+yis minimal
when x=y=1.
Ifxy=1 ,t h e n y=1
x. It follows that x+y=x+1
x=/parenleftBig√
x−1
√
x/parenrightBig2
+2 ,
and then x+yis minimal when√
x−1
√
x=0 ,t h a ti s ,w h e n x= 1. Therefore,
x=y=1 .
This can also be interpreted geometrically in the following way, “of all the
rectangles with area 1, the square has the smallest perimeter” .I nf a c t ,i f x,yare
the lengths of the sides of the rectangle, its area is xy= 1 and its perimeter is
2(x+y)=2/parenleftbig
x+1
x/parenrightbig
=2/braceleftbigg/parenleftBig√
x−1
√
x/parenrightBig2
+2/bracerightbigg
≥4. Moreover, the perimeter is 4 if
and only if√
x−1
√
x=0 ,t h a ti s ,w h e n x=y=1 .
Example 1.2.3. For any positive number x, we have x+1
x≥2.

6 Numerical Inequalities
Observe that x+1
x=/parenleftBig√
x−1
√
x/parenrightBig2
+2≥2.Moreover, the equality holds if
and only if√
x−1
√
x=0 ,t h a ti s ,w h e n x=1.
Example 1.2.4. Ifa,b>0,t h e na
b+b
a≥2, and the equality holds if and only if
a=b.
It is enough to consider the previous example with x=a
b.
Example 1.2.5. Given a,b,c>0, it is possible to construct a triangle with sides
of length a,b,cif and only if pa2+qb2>p q c2for any p,qwithp+q=1.
Remember that a,bandcare the lengths of the sides of a triangle if and
only if a+b>c,a+c>b andb+c>a.
Let
Q=pa2+qb2−pqc2=pa2+( 1−p)b2−p(1−p)c2=c2p2+(a2−b2−c2)p+b2,
therefore Qis a quadratic function1inpand
Q>0⇔/triangle=/bracketleftBig/parenleftbig
a2−b2−c2/parenrightbig2−4b2c2/bracketrightBig
<0
⇔/bracketleftbig
a2−b2−c2−2bc/bracketrightbig/bracketleftbig
a2−b2−c2+2bc/bracketrightbig
<0
⇔/bracketleftbig
a2−(b+c)2/bracketrightbig/bracketleftbig
a2−(b−c)2/bracketrightbig
<0
⇔[a+b+c][a−b−c][a−b+c][a+b−c]<0
⇔[b+c−a][c+a−b][a+b−c]>0.
Now, [ b+c−a][c+a−b][a+b−c]>0 if the three factors are positive or if one of
them is positive and the other two are negative. However, the latter is impossible,
because if [ b+c−a]<0a n d[ c+a−b]<0, we would have, adding these two
inequalities, that c<0, which is false. Therefore the three factors are necessarily
positive.
Exercise 1.16. Suppose the polynomial ax2+bx+csatisﬁes the following: a>0,
a+b+c≥0,a−b+c≥0,a−c≥0a n d b2−4ac≥0. Prove that the roots are
real and that they belong to the interval −1≤x≤1.
Exercise 1.17. Ifa,b,care positive numbers, prove that it is not possible for the
inequalities a(1−b)>1
4,b(1−c)>1
4,c(1−a)>1
4to hold at the same time.
1A quadratic function ax2+bx+cwitha>0 is positive when its discriminant Δ = b2−4ac
is negative, in fact, this follows from ax2+bx+c=a(x+b
2a)2+4ac−b2
4a. Remember that the
roots are−b±√
b2−4ac
2a, and they are real when Δ ≥0, otherwise they are not real roots, and
thenax2+bx+cwill have the same sign; this expression will be positive if a>0.

1.3 Arithmetic mean-geometric mean 7
1.3 A fundamental inequality,
arithmetic mean-geometric mean
The ﬁrst inequality that we consider, fundamental in optimization problems, is
the inequality between the arithmetic mean and the geometric mean of two non-
negative numbers aandb, which is expressed as
a+b
2≥√
ab, (AM-GM).
Moreover, the equality holds if and only if a=b.
The numbersa+b
2and√
abare known as the arithmetic mean and the ge-
ometric mean ofaandb, respectively. To prove the inequality we only need to
observe that
a+b
2−√
ab=a+b−2√
ab
2=1
2/parenleftBig√
a−√
b/parenrightBig2
≥0.
And the equality holds if and only if√
a=√
b,t h a ti s ,w h e na =b.
Exercise 1.18. Forx≥0, prove that 1 + x≥2√
x.
Exercise 1.19. Forx>0, prove that x+1
x≥2.
Exercise 1.20. Forx,y∈R+,p r o v et h a t x2+y2≥2xy.
Exercise 1.21. Forx,y∈R+, prove that 2( x2+y2)≥(x+y)2.
Exercise 1.22. Forx,y∈R+,p r o v et h a t1
x+1
y≥4
x+y.
Exercise 1.23. Fora,b,x∈R+,p r o v et h a t ax+b
x≥2√
ab.
Exercise 1.24. Ifa,b>0, thena
b+b
a≥2.
Exercise 1.25. If 0<b≤a,t h e n1
8(a−b)2
a≤a+b
2−√
ab≤1
8(a−b)2
b.
Now, we will present a geometric and a visual proof of the following inequal-
ities, for x,y>0,
2
1
x+1
y≤√
xy≤x+y
2. (1.1)
x yhgA
BCDE
O

8 Numerical Inequalities
Letx=BD,y=DCand let us construct a semicircle of diameter BC=
x+y.L e t Abe the point where the perpendicular to BCinDintersects the
semicircle and let Ebe the perpendicular projection from Dto the radius AO.
Let us write AD=handAE=g.S i n c e ABD andCAD are similar right triangles,
we deduce thath
y=x
h,t h e n h=√
xy.
Also, since AOD andADE are similar right triangles, we have
g
√
xy=√
xy
x+y
2,t h e n g=2xy
x+y=2
/parenleftBig
1
x+1
y/parenrightBig.
Finally, the geometry tells us that in a right triangle, the length of one leg is
always smaller than the length of the hypotenuse. Hence, g≤h≤x+y
2,w h i c hc a n
be written as2
1
x+1
y≤√
xy≤x+y
2.
The number2
1
x+1
yis known as the harmonic mean ofxandy, and the left inequality
in (1.1) is known as the inequality between the harmonic mean and the geometric
mean.
Some inequalities can be proved through the multiple application of a simple
inequality and the use of a good idea to separate the problem into parts that are
easier to deal with, a method which is often used to solve the following exercises.
Exercise 1.26. Forx,y,z∈R+,(x+y)(y+z)(z+x)≥8xyz.
Exercise 1.27. Forx,y,z∈R,x2+y2+z2≥xy+yz+zx.
Exercise 1.28. Forx,y,z∈R+,xy+yz+zx≥x√
yz+y√
zx+z√
xy.
Exercise 1.29. Forx,y∈R,x2+y2+1≥xy+x+y.
Exercise 1.30. Forx,y,z∈R+,1
x+1
y+1
z≥1
√
xy+1
√
yz+1
√
zx.
Exercise 1.31. Forx,y,z∈R+,xy
z+yz
x+zx
y≥x+y+z.
Exercise 1.32. Forx,y,z∈R,x2+y2+z2≥x/radicalbig
y2+z2+y√
x2+z2.
The inequality between the arithmetic mean and the geometric mean can
be extended to more numbers. For instance, we can prove the following inequa-
lity between the arithmetic mean and the geometric mean of four non-negative
numbers a,b,c,d, expressed asa+b+c+d
4≥4√
abcd, in the following way:
a+b+c+d
4=1
2/parenleftbigga+b
2+c+d
2/parenrightbigg
≥1
2/parenleftBig√
ab+√
cd/parenrightBig
≥/radicalBig
√
ab√
cd=4√
abcd.

1.3 Arithmetic mean-geometric mean 9
Observe that we have used the AM-GM inequality three times for two numbers
in each case: with aandb,w i t h candd,a n dw i t h√
aband√
cd.M o r e o v e r ,t h e
equality holds if and only if a=b,c=dandab=cd, that is, when the numbers
satisfy a=b=c=d.
Exercise 1.33. Forx,y∈R,x4+y4+8≥8xy.
Exercise 1.34. Fora,b,c,d∈R+,(a+b+c+d)/parenleftbig1
a+1
b+1
c+1
d/parenrightbig
≥16.
Exercise 1.35. Fora,b,c,d∈R+,a
b+b
c+c
d+d
a≥4.
A useful trick also exists for checking that the inequalitya+b+c
3≥3√
abc
is true for any three non-negative numbers a,bandc. Consider the following
four numbers a,b,candd=3√
abc. Since the AM-GM inequality holds for four
numbers, we havea+b+c+d
4≥4√
abcd=4√
d3d=d.T h e na+b+c
4≥d−1
4d=3
4d.
Hence,a+b+c
3≥d=3√
abc.
These ideas can be used to justify the general version of the inequality for n
non-negative numbers. If a1,a2,…,a narennon-negative numbers, we take the
numbers AandGas
A=a1+a2+···+an
nand G=n√
a1a2···an.
T h e s en u m b e r sa r ek n o w na st h e arithmetic mean and the geometric mean of the
numbers a1,a2,…,a n, respectively.
Theorem 1.3.1 (The AM-GM inequality).
a1+a2+···+an
n≥n√
a1a2···an.
First proof (Cauchy). LetPnbe the statement G≤A,f o rnnumbers. We will
proceed by mathematical induction on n, but this is an induction of the following
type.
(1) We prove that the statement is true for 2 numbers, that is, P2is true.
(2) We prove that Pn⇒Pn−1.
(3) We prove that Pn⇒P2n.
When (1), (2) and (3) are veriﬁed, all the assertions Pnwithn≥2a r es h o w n
to be true. Now, we will prove these statements.
(1) This has already been done in the ﬁrst part of the section.
(2) Let a1,…,an−1be non-negative numbers and let g=n−1√
a1···an−1.U s i n g
this number and the numbers we already have, i.e., a1,…,an−1,w eg e t n
numbers to which we apply Pn,
a1+···+an−1+g
n≥n√
a1a2···an−1g=n/radicalbig
gn−1·g=g.

10 Numerical Inequalities
We deduce that a1+···+an−1+g≥ng, and then it follows thata1+···+an−1
n−1≥
g, therefore Pn−1is true.
(3) Let a1,a2,…,a2nbe non-negative numbers, then
a1+a2+···+a2n=(a1+a2)+(a3+a4)+···+(a2n−1+a2n)
≥2/parenleftbig√
a1a2+√
a3a4+···+√
a2n−1a2n/parenrightbig
≥2n/parenleftbig√
a1a2√
a3a4···√
a2n−1a2n/parenrightbig1
n
=2n(a1a2···a2n)1
2n.
We have applied the statement P2several times, and we have also applied the
statement Pnto the numbers√
a1a2,√
a3a4,…,√
a2n−1a2n. /square
Second proof. LetA=a1+···+an
n.W et a k et w on u m b e r s ai, one smaller than A
and the other greater than A(if they exist), say a1=A−handa2=A+k,w i t h
h,k>0.
We exchange a1anda2for two numbers that increase the product and ﬁx
the sum, deﬁned as
a/prime
1=A,a/prime2=A+k−h.
Since a/prime
1+a/prime2=A+A+k−h=A−h+A+k=a1+a2, clearly a/prime1+a/prime2+a3+
···+an=a1+a2+a3+···+an, buta/prime
1a/prime2=A(A+k−h)=A2+A(k−h)a n d
a1a2=(A+k)(A−h)=A2+A(k−h)−hk,t h e n a/prime
1a/prime2>a 1a2and thus it follows
thata/prime1a/prime2a3···an>a 1a2a3···an.
IfA=a/prime1=a/prime2=a3=···=an, there is nothing left to prove (the equality
holds), otherwise two elements will exist, one greater than Aand the other one
smaller than Aand the argument is repeated. Since every time we perform this
operation we create a number equal to A, this process can not be used more than
ntimes. /square
Example 1.3.2. Find the maximum value of x(1−x3)for0≤x≤1.
The idea of the proof is to exchange the product for another one in such
a way that the sum of the elements involved in the new product is constant. If
y=x(1−x3), it is clear that the right side of 3 y3=3×3(1−x3)(1−x3)(1−x3),
expressed as the product of four numbers 3 x3,( 1−x3), (1−x3)a n d( 1 −x3), has
a constant sum equal to 3. The AM-GM inequality for four numbers tells us that
3y3≤/parenleftbigg3x3+3 ( 1−x3)
4/parenrightbigg4
=/parenleftbigg3
4/parenrightbigg4
.
Thus y≤3
43√
4. Moreover, the maximum value is reached using 3 x3=1−x3,t h a t
is, ifx=1
3√
4.

1.3 Arithmetic mean-geometric mean 11
Exercise 1.36. Letxi>0,i=1,…,n .P r o v et h a t
(x1+x2+···+xn)/parenleftbigg1
x1+1
x2+···+1
xn/parenrightbigg
≥n2.
Exercise 1.37. If{a1,…,a n}is a permutation of {b1,…,b n}⊂R+,t h e n
a1
b1+a2
b2+···+an
bn≥nandb1
a1+b2
a2+···+bn
an≥n.
Exercise 1.38. Ifa>1, then an−1>n/parenleftBig
an+1
2−an−1
2/parenrightBig
.
Exercise 1.39. Ifa,b,c>0a n d( 1+ a)(1 +b)(1 +c)=8 ,t h e n abc≤1.
Exercise 1.40. Ifa,b,c>0, thena3
b+b3
c+c3
a≥ab+bc+ca.
Exercise 1.41. For non-negative real numbers a,b,c,p r o v et h a t
a2b2+b2c2+c2a2≥abc(a+b+c).
Exercise 1.42. Ifa,b,c>0, then
/parenleftbig
a2b+b2c+c2a/parenrightbig/parenleftbig
ab2+bc2+ca2/parenrightbig
≥9a2b2c2.
Exercise 1.43. Ifa,b,c>0 satisfy that abc=1 ,p r o v et h a t
1+ab
1+a+1+bc
1+b+1+ac
1+c≥3.
Exercise 1.44. Ifa,b,c>0, prove that
1
a+1
b+1
c≥2/parenleftbigg1
a+b+1
b+c+1
c+a/parenrightbigg
≥9
a+b+c.
Exercise 1.45. IfHn=1+1
2+···+1
n,p r o v et h a t
n(n+1 )1
n<n+Hnforn≥2.
Exercise 1.46. Letx1,x2,…,xn>0 such that1
1+x1+···+1
1+xn=1 .P r o v et h a t
x1x2···xn≥(n−1)n.
Exercise 1.47. (Short list IMO, 1998) Let a1,a2,…,anbe positive numbers with
a1+a2+···+an<1, prove that
a1a2···an[1−(a1+a2+···+an)]
(a1+a2+···+an)( 1−a1)( 1−a2)···(1−an)≤1
nn+1.

12 Numerical Inequalities
Exercise 1.48. Leta1,a2,…,anbe positive numbers such that1
1+a1+···+1
1+an=
1.Prove that
√
a1+···+√
an≥(n−1)/parenleftbigg1
√
a1+···+1
√
an/parenrightbigg
.
Exercise 1.49. (APMO, 1991) Let a1,a2,…,an,b1,b2,…,bnbe positive numbers
witha1+a2+···+an=b1+b2+···+bn.Prove that
a2
1
a1+b1+···+a2n
an+bn≥1
2(a1+···+an).
Exercise 1.50. Leta,b,cbe positive numbers, prove that
1
a3+b3+abc+1
b3+c3+abc+1
c3+a3+abc≤1
abc.
Exercise 1.51. Leta,b,cbe positive numbers with a+b+c=1 ,p r o v et h a t
/parenleftbigg1
a+1/parenrightbigg/parenleftbigg1
b+1/parenrightbigg/parenleftbigg1
c+1/parenrightbigg
≥64.
Exercise 1.52. Leta,b,cbe positive numbers with a+b+c=1 ,p r o v et h a t
/parenleftbigg1
a−1/parenrightbigg/parenleftbigg1
b−1/parenrightbigg/parenleftbigg1
c−1/parenrightbigg
≥8.
Exercise 1.53. (Czech and Slovak Republics, 2005) Let a,b,cbe positive numbers
that satisfy abc=1 ,p r o v et h a t
a
(a+1 ) (b+1 )+b
(b+1 ) (c+1 )+c
(c+1 ) (a+1 )≥3
4.
Exercise 1.54. Leta,b,cbe positive numbers for which1
1+a+1
1+b+1
1+c=1 .
Prove that
abc≥8.
Exercise 1.55. Leta,b,cbe positive numbers, prove that
2ab
a+b+2bc
b+c+2ca
c+a≤a+b+c.
Exercise 1.56. Leta1,a2,…,an,b1,b2,…,bnbe positive numbers, prove that
n/summationdisplay
i=11
aibin/summationdisplay
i=1(ai+bi)2≥4n2.
Exercise 1.57. (Russia, 1991) For all non-negative real numbers x,y,z,p r o v et h a t
(x+y+z)2
3≥x√
yz+y√
zx+z√
xy.

1.4 A wonderful inequality 13
Exercise 1.58. (Russia, 1992) For all positive real numbers x,y,z,p r o v et h a t
x4+y4+z2≥√
8xyz.
Exercise 1.59. (Russia, 1992) For any real numbers x,y>1, prove that
x2
y−1+y2
x−1≥8.
1.4 A wonderful inequality:
The rearrangement inequality
Consider two collections of real numbers in increasing order,
a1≤a2≤···≤ anand b1≤b2≤···≤ bn.
For any permutation ( a/prime
1,a/prime2,…,a/primen)o f(a1,a2,…,a n), it happens that
a1b1+a2b2+···+anbn≥a/prime
1b1+a/prime2b2+···+a/primenbn (1.2)
≥anb1+an−1b2+···+a1bn. (1.3)
Moreover, the equality in (1.2) holds if and only if ( a/prime
1,a/prime2,…,a/primen)=(a1,a2,…,a n).
And the equality in (1.3) holds if and only if ( a/prime
1,a/prime2,…,a/primen)=(an,an−1,…,a 1).
Inequality (1.2) is known as the rearrangement inequality .
Corollary 1.4.1. For any permutation (a/prime1,a/prime2,…,a/primen)of(a1,a2,…,a n),i tf o l l o w s
that
a2
1+a22+···+a2n≥a1a/prime1+a2a/prime2+···+ana/primen.
Corollary 1.4.2. For any permutation (a/prime
1,a/prime2,…,a/primen)of(a1,a2,…,a n),i tf o l l o w s
that
a/prime
1
a1+a/prime2
a2+···+a/primen
an≥n.
Proof (of the rearrangement inequality). Suppose that b1≤b2≤···≤ bn.Let
S=a1b1+a2b2+···+arbr+···+asbs+···+anbn,
S/prime=a1b1+a2b2+···+asbr+···+arbs+···+anbn.
The diﬀerence between SandS/primeis that the coeﬃcients of brandbs,w h e r e r<s,
are switched. Hence
S−S/prime=arbr+asbs−asbr−arbs=(bs−br)(as−ar).
Thus, we have that S≥S/primeif and only if as≥ar. Repeating this process we get
the result that the sum Sis maximal when a1≤a2≤···≤ an. /square

14 Numerical Inequalities
Example 1.4.3. (IMO, 1975) Consider two collections of numbers x1≤x2≤···≤
xnandy1≤y2≤···≤ yn, and one permutation (z1,z2,…,z n)of(y1,y2,…,y n).
Prove that
(x1−y1)2+···+(xn−yn)2≤(x1−z1)2+···+(xn−zn)2.
By squaring and rearranging this last inequality, we ﬁnd that it is equivalent
to
n/summationdisplay
i=1×2
i−2n/summationdisplay
i=1xiyi+n/summationdisplay
i=1y2
i≤n/summationdisplay
i=1x2i−2n/summationdisplay
i=1xizi+n/summationdisplay
i=1z2
i,
but since/summationtextn
i=1y2
i=/summationtextni=1z2
i, then the inequality we have to prove turns to be
equivalent to
n/summationdisplay
i=1xizi≤n/summationdisplay
i=1xiyi,
which in turn is inequality (1.2).
Example 1.4.4. (IMO, 1978) Letx1,x2,…,xnbe distinct positive integers, prove
that
x1
12+x2
22+···+xn
n2≥1
1+1
2+···+1
n.
Let (a1,a2,…,a n)b eap e r m u t a t i o no f( x1,x2,…,x n)w i t ha 1≤a2≤
···≤ anand let ( b1,b2,…,b n)=/parenleftBig
1
n2,1
(n−1)2,…,1
12/parenrightBig
;t h a ti s , bi=1
(n+1−i)2for
i=1,…,n.
Consider the permutation ( a/prime
1,a/prime2,…,a/primen)o f(a1,a2,…,a n) deﬁned by a/prime
i=
xn+1−i,f o ri=1,…,n . Using inequality (1.3) we can argue that
x1
12+x2
22+···+xn
n2=a/prime
1b1+a/prime2b2+···+a/primenbn
≥anb1+an−1b2+···+a1bn
=a1bn+a2bn−1+···+anb1
=a1
12+a2
22+···+an
n2.
Since 1 ≤a1,2≤a2,…,n≤an,w eh a v et h a t
x1
12+x2
22+···+xn
n2≥a1
12+a2
22+···+an
n2≥1
12+2
22+···+n
n2=1
1+1
2+···+1
n.
Example 1.4.5. (IMO, 1964) Suppose that a,b,care the lengths of the sides of a
triangle. Prove that
a2(b+c−a)+b2(a+c−b)+c2(a+b−c)≤3abc.

1.4 A wonderful inequality 15
Since the expression is a symmetric function of a,bandc, we can as-
sume, without loss of generality, that c≤b≤a.I nt h i sc a s e , a(b+c−a)≤
b(a+c−b)≤c(a+b−c).
For instance, the ﬁrst inequality is proved in the following way:
a(b+c−a)≤b(a+c−b)⇔ab+ac−a2≤ab+bc−b2
⇔(a−b)c≤(a+b)(a−b)
⇔(a−b)(a+b−c)≥0.
By (1.3) of the rearrangement inequality, we have
a2(b+c−a)+b2(c+a−b)+c2(a+b−c)≤ba(b+c−a)+cb(c+a−b)+ac(a+b−c),
a2(b+c−a)+b2(c+a−b)+c2(a+b−c)≤ca(b+c−a)+ab(c+a−b)+bc(a+b−c).
Therefore, 2/bracketleftbig
a2(b+c−a)+b2(c+a−b)+c2(a+b−c)/bracketrightbig
≤6abc.
Example 1.4.6. (IMO, 1983) Leta,bandcbe the lengths of the sides of a triangle.
Prove that
a2b(a−b)+b2c(b−c)+c2a(c−a)≥0.
Consider the case c≤b≤a(the other cases are similar).
As in the previous example, we have that a(b+c−a)≤b(a+c−b)≤c(a+b−c)
and since1
a≤1
b≤1
c, using Inequality (1.2) leads us to
1
aa(b+c−a)+1
bb(c+a−b)+1
cc(a+b−c)
≥1
ca(b+c−a)+1
ab(c+a−b)+1
bc(a+b−c).
Therefore,
a+b+c≥a(b−a)
c+b(c−b)
a+c(a−c)
b+a+b+c.
It follows thata(b−a)
c+b(c−b)
a+c(a−c)
b≤0.Multiplying by abc,w eo b t a i n
a2b(a−b)+b2c(b−c)+c2a(c−a)≥0.
Example 1.4.7 (Cauchy-Schwarz inequality). For real numbers x1,…,xn,y1,…,
yn, the following inequality holds:
/parenleftBiggn/summationdisplay
i=1xiyi/parenrightBigg2
≤/parenleftBiggn/summationdisplay
i=1×2
i/parenrightBigg/parenleftBiggn/summationdisplay
i=1y2
i/parenrightBigg
.
The equality holds if and only if there exists some λ∈Rwithxi=λyifor all
i=1,2,…,n .

16 Numerical Inequalities
Ifx1=x2=···=xn=0o r y1=y2=···=yn= 0, the result is evident.
Otherwise, let S=/radicalbig
/summationtextn
i=1×2
iandT=/radicalbig
/summationtextn
i=1y2
i, where it is clear that S,T/negationslash=0 .
Takeai=xi
Sandan+i=yi
Tfori=1,2,…,n. Using Corollary 1.4.1,
2=n/summationdisplay
i=1×2
i
S2+n/summationdisplay
i=1y2
i
T2=2n/summationdisplay
i=1a2
i
≥a1an+1+a2an+2+···+ana2n+an+1a1+···+a2nan
=2x1y1+x2y2+···+xnyn
ST.
The equality holds if and only if ai=an+ifori=1,2,…,n , or equivalently, if
and only if xi=S
Tyifori=1,2,…,n .
Another proof of the Cauchy-Schwarz inequality can be established using
Lagrange’s identity
/parenleftBiggn/summationdisplay
i=1xiyi/parenrightBigg2
=n/summationdisplay
i=1x2in
/summationdisplay
i=1y2
i−1
2n/summationdisplay
i=1n/summationdisplay
j=1(xiyj−xjyi)2.
The importance of the Cauchy-Schwarz inequality will be felt throughout the
remaining part of this book, as we will use it as a tool to solve many exercises and
problems proposed here.
Example 1.4.8 (Nesbitt’s inequality). Fora,b,c∈R+, we have
a
b+c+b
c+a+c
a+b≥3
2.
Without loss of generality, we can assume that a≤b≤c, and then it follows
thata+b≤c+a≤b+cand1
b+c≤1
c+a≤1
a+b.
Using the rearrangement inequality (1.2) twice, we obtain
a
b+c+b
c+a+c
a+b≥b
b+c+c
c+a+a
a+b,
a
b+c+b
c+a+c
a+b≥c
b+c+a
c+a+b
a+b.
Hence,
2/parenleftbigga
b+c+b
c+a+c
a+b/parenrightbigg
≥/parenleftbiggb+c
b+c+c+a
c+a+a+b
a+b/parenrightbigg
=3.
Another way to prove the inequality is using Inequality (1.3) twice,
c+a
b+c+a+b
c+a+b+c
a+b≥3,
a+b
b+c+b+c
c+a+c+a
a+b≥3.

1.4 A wonderful inequality 17
Then, after adding the two expressions, we get2a+b+c
b+c+2b+c+a
c+a+2c+a+b
a+b≥6,
therefore
2a
b+c+2b
c+a+2c
a+b≥3.
Example 1.4.9. (IMO, 1995) Leta,b,cbe positive real numbers with abc=1.
Prove that
1
a3(b+c)+1
b3(c+a)+1
c3(a+b)≥3
2.
Without loss of generality, we can assume that c≤b≤a.L e tx=1
a,y=1
b
andz=1
c,t h u s
S=1
a3(b+c)+1
b3(c+a)+1
c3(a+b)
=x3
1
y+1
z+y3
1
z+1
x+z3
1
x+1
y
=x2
y+z+y2
z+x+z2
x+y.
Since x≤y≤z, we can deduce that x+y≤z+x≤y+zand also that
x
y+z≤y
z+x≤z
x+y. Using the rearrangement inequality (1.2), we show that
x2
y+z+y2
z+x+z2
x+y≥xy
y+z+yz
z+x+zx
x+y,
x2
y+z+y2
z+x+z2
x+y≥xz
y+z+yx
z+x+zy
x+y,
w h i c hi nt u r nl e a d st o2 S≥x+y+z≥33√
xyz= 3. Therefore, S≥3
2.
Example 1.4.10. (APMO, 1998) Leta,b,c∈R+, prove that
/parenleftBig
1+a
b/parenrightBig/parenleftbigg
1+b
c/parenrightbigg/parenleftBig
1+c
a/parenrightBig
≥2/parenleftbigg
1+a+b+c
3√
abc/parenrightbigg
.
Observe that
/parenleftBig
1+a
b/parenrightBig/parenleftbigg
1+b
c/parenrightbigg/parenleftBig
1+c
a/parenrightBig
≥2/parenleftbigg
1+a+b+c
3√
abc/parenrightbigg
⇔1+/parenleftbigga
b+b
c+c
a/parenrightbigg
+/parenleftbigga
c+c
b+b
a/parenrightbigg
+abc
abc≥2/parenleftbigg
1+a+b+c
3√
abc/parenrightbigg
⇔a
b+b
c+c
a+a
c+c
b+b
a≥2(a+b+c)
3√
abc.

18 Numerical Inequalities
Now we set a=x3,b=y3,c=z3. We need to prove that
x3
y3+y3
z3+z3
x3+x3
z3+z3
y3+y3
x3≥2/parenleftbig
x3+y3+z3/parenrightbig
xyz.
But, if we consider
(a1,a2,a3,a4,a5,a6)=/parenleftbiggx
y,y
z,z
x,x
z,z
y,y
x/parenrightbigg
,
(a/prime
1,a/prime2,a/prime3,a/prime4,a/prime5,a/prime6)=/parenleftbiggy
z,z
x,x
y,z
y,y
x,x
z/parenrightbigg
,
(b1,b2,b3,b4,b5,b6)=/parenleftbiggx2
y2,y2
z2,z2
x2,x2
z2,z2
y2,y2
x2/parenrightbigg
,
we are led to the following result:
x3
y3+y3
z3+z3
x3+x3
z3+z3
y3+y3
x3≥x2
y2y
z+y2
z2z
x+z2
x2x
y+x2
z2z
y+z2
y2y
x+y2
x2x
z
=x2
yz+y2
zx+z2
xy+x2
zy+z2
yx+y2
xz
=2/parenleftbig
x3+y3+z3/parenrightbig
xyz.
Example 1.4.11 (Tchebyshev’s inequality). Leta1≤a2≤···≤ anandb1≤b2≤
···≤bn,t h e n
a1b1+a2b2+···+anbn
n≥a1+a2+···+an
n·b1+b2+···+bn
n.
Applying the rearrangement inequality several times, we get
a1b1+···+anbn=a1b1+a2b2+···+anbn,
a1b1+···+anbn≥a1b2+a2b3+···+anb1,
a1b1+···+anbn≥a1b3+a2b4+···+anb2,
………
a1b1+···+anbn≥a1bn+a2b1+···+anbn−1,
and adding together all the expressions, we obtain
n(a1b1+···+anbn)≥(a1+···+an)(b1+···+bn).
The equality holds when a1=a2=···=anorb1=b2=···=bn.
Exercise 1.60. Any three positive real numbers a,bandcsatisfy the following
inequality:
a3+b3+c3≥a2b+b2c+c2a.

1.4 A wonderful inequality 19
Exercise 1.61. Any three positive real numbers a,bandc,w i t h abc=1 ,s a t i s f y
a3+b3+c3+(ab)3+(bc)3+(ca)3≥2(a2b+b2c+c2a).
Exercise 1.62. Any three positive real numbers a,bandcsatisfy
a2
b2+b2
c2+c2
a2≥b
a+c
b+a
c.
Exercise 1.63. Any three positive real numbers a,bandcsatisfy
1
a2+1
b2+1
c2≥a+b+c
abc.
Exercise 1.64. Ifa,bandcare the lengths of the sides of a triangle, prove that
a
b+c−a+b
c+a−b+c
a+b−c≥3.
Exercise 1.65. Ifa1,a2,…,an∈R+ands=a1+a2+···+an,t h e n
a1
s−a1+a2
s−a2+···+an
s−an≥n
n−1.
Exercise 1.66. Ifa1,a2,…,an∈R+ands=a1+a2+···+an,t h e n
s
s−a1+s
s−a2+···+s
s−an≥n2
n−1.
Exercise 1.67. Ifa1,a2,…,an∈R+anda1+a2+···+an=1 ,t h e n
a1
2−a1+a2
2−a2+···+an
2−an≥n
2n−1.
Exercise 1.68. (Quadratic mean-arithmetic mean inequality) Letx1,…,xn∈
R+,t h e n/radicalbigg
x2
1+x22+···+x2n
n≥x1+x2+···+xn
n.
Exercise 1.69. For positive real numbers a,b,csuch that a+b+c=1 ,p r o v et h a t
ab+bc+ca≤1
3.
Exercise 1.70. (Harmonic, geometric and arithmetic mean) Letx1,…,xn∈R+,
prove that
n
1
x1+1
x2+···+1
xn≤n√
x1x2···xn≤x1+x2+···+xn
n.
And the equalities hold if and only if x1=x2=···=xn.

20 Numerical Inequalities
Exercise 1.71. Leta1,a2,…,anbe positive numbers with a1a2···an=1 .P r o v e
that
an−1
1+an−1
2+···+an−1
n≥1
a1+1
a2+···+1
an.
Exercise 1.72. (China, 1989) Let a1,a2,…,anbe positive numbers such that
a1+a2+···+an=1 .P r o v et h a t
a1
√
1−a1+···+an
√
1−an≥1
√
n−1(√
a1+···+√
an).
Exercise 1.73. Leta,bandcbe positive numbers such that a+b+c=1 .P r o v e
that
(i)√
4a+1+√
4b+1+√
4c+1<5,
(ii)√
4a+1+√
4b+1+√
4c+1≤√
21.
Exercise 1.74. Leta,b,c,d∈R+withab+bc+cd+da=1 ,p r o v et h a t
a3
b+c+d+b3
a+c+d+c3
a+b+d+d3
a+b+c≥1
3.
Exercise 1.75. Leta,b,cbe positive numbers with abc=1 ,p r o v et h a t
a
b+b
c+c
a≥a+b+c.
Exercise 1.76. Letx1,x2,…,xn(n>2) be real numbers such that the sum of
anyn−1 of them is greater than the element left out of the sum. Set s=/summationtextn
k=1xk.
Prove that
n/summationdisplay
k=1×2
k
s−2xk≥s
n−2.
1.5 Convex functions
A function f:[a,b]→Ris called convex in the interval I=[a,b] if for any
t∈[0,1] and for all a≤x<y ≤b, the following inequality holds:
f(ty+( 1− t)x)≤tf(y)+( 1 −t)f(x). (1.4)
Geometrically, the inequality in the deﬁnition means that the graph of f
between xandyis below the segment which joins the points (x, f (x)) and ( y,f(y)).

1.5 Convex functions 21
x y(x, f(x))
(y,f(y))
In fact, the equation of the line joining the points ( x, f(x)) and ( y,f(y)) is
expressed as
L(s)=f(x)+f(y)−f(x)
y−x(s−x).
Then, evaluating at the point s=ty+( 1−t)x,w eg e t
L(ty+( 1−t)x)=f(x)+f(y)−f(x)
y−x(t(y−x)) =f(x)+t(f(y)−f(x))
=tf(y)+( 1 −t)f(x).
Hence, Inequality (1.4) is equivalent to
f(ty+( 1−t)x)≤L(ty+( 1−t)x).
Proposition 1.5.1. (1)Iffis convex in the interval [a,b], then it is convex in
any subinterval [x, y]⊂[a,b].
(2)Iffis convex in [a,b], then for any x,y∈[a,b], we have that
f/parenleftbiggx+y
2/parenrightbigg
≤1
2(f(x)+f(y)). (1.5)
(3)(Jensen’s inequality) Iffis convex in [a,b], then for any t1,…,tn∈[0,1],
with/summationtextn
i=1ti=1,a n df o r x1,…,xn∈[a,b], we can deduce that
f(t1x1+···+tnxn)≤t1f(x1)+···+tnf(xn).
(4)In particular, for x1,…,xn∈[a,b], we can establish that
f/parenleftbiggx1+···+xn
n/parenrightbigg
≤1
n(f(x1)+···+f(xn)) .

22 Numerical Inequalities
Proof. (1) We leave the proof as an exercise for the reader.
( 2 )I ti ss u ﬃ c i e n tt oc h o o s e t=1
2in (1.4).
(3) We have
f(t1x1+···+tnxn)=f((1−tn)(t1
1−tnx1+···+tn−1
1−tnxn−1)+tnxn)
≤(1−tn)f/parenleftbiggt1
1−tnx1+···+tn−1
1−tnxn−1/parenrightbigg
+tnf(xn), by convexity
≤(1−tn)/braceleftbiggt1
1−tnf(x1)+···+tn−1
1−tnf(xn−1)/bracerightbigg
+tnf(xn), by induction
=t1f(x1)+···+tnf(xn).
(4) We only need to apply (3) using t1=t2=···=tn=1
n. /square
Observations 1.5.2. (i)We can see that (4)holds true only under the assumption
thatfsatisﬁes the relation f/parenleftbigx+y
2/parenrightbig
≤f(x)+f(y)
2for any x,y∈[a,b].
(ii)We can observe that (3)is true for t1,…,tn∈[0,1]rational numbers, only
under the condition that fsatisﬁes the relation f/parenleftbigx+y
2/parenrightbig
≤f(x)+f(y)
2for any
x,y∈[a,b].
We will prove (i) using induction. Let us call Pnthe assertion
f/parenleftbiggx1+···+xn
n/parenrightbigg
≤1
n(f(x1)+···+f(xn))
forx1,…,xn∈[a,b]. It is clear that P1andP2are true.
Now, we will show that Pn⇒Pn−1.
Letx1,…,xn∈[a,b]a n dl e t y=x1+···+xn−1
n−1.S i n c e Pnis true, we can
establish that
f/parenleftbiggx1+···+xn−1+y
n/parenrightbigg
≤1
nf(x1)+···+1
nf(xn−1)+1
nf(y).
But the left side is f(y), therefore n·f(y)≤f(x1)+···+f(xn−1)+f(y), and
f(y)≤1
n−1(f(x1)+···+f(xn−1)).
Finally, we can observe that Pn⇒P2n.
LetD=f/parenleftBig
x1+···+xn+xn+1+···+x2n
2n/parenrightBig
=f/parenleftbigu+v
2/parenrightbig
,w h e r e u=x1+···+xn
nand
v=xn+1+···+x2n
n.
Since f/parenleftbigu+v
2/parenrightbig
≤1
2(f(u)+f(v)), we have that
D≤1
2(f(u)+f(v)) =1
2/parenleftbigg
f/parenleftbiggx1+···+xn
n/parenrightbigg
+f/parenleftbiggxn+1+···+x2n
n/parenrightbigg/parenrightbigg
≤1
2n(f(x1)+···+f(xn)+f(xn+1)+···+f(x2n)) ,

1.5 Convex functions 23
where we have used twice the statement that Pnis true.
To prove (ii), our starting point will be the assertion that f/parenleftbigx1+···+xn
n/parenrightbig
≤
1
n(f(x1)+···+f(xn)) for x1,…,xn∈[a,b]a n d n∈N.
Lett1=r1
s1,…,tn=rn
snbe rational numbers in [0, 1] with/summationtextn
i=1ti=1 .
Ifmis the least common multiple of the si’s, then ti=pi
mwithpi∈Nand/summationtextni=1pi=m, hence
f(t1x1+···+tnxn)=f/parenleftBigp1
mx1+···+pn
mxn/parenrightBig
=f⎛
⎜⎝1
m⎡
⎢⎣(x1+···+x1)/bracehtipupleft
/bracehtipdownright/bracehtipdownleft
/bracehtipupright
p1−terms+···+(xn+···+xn)/bracehtipupleft
/bracehtipdownright/bracehtipdownleft
/bracehtipupright
pn−terms⎤
⎥⎦⎞
⎟⎠
≤1
m⎡
⎢⎣(f(x1)+···+f(x1))/bracehtipupleft
/bracehtipdownright/bracehtipdownleft
/bracehtipupright
p1−terms+···+(f(xn)+···+f(xn))/bracehtipupleft
/bracehtipdownright/bracehtipdownleft
/bracehtipupright
pn−terms⎤
⎥⎦
=p1
mf(x1)+···+pn
mf(xn)
=t1f(x1)+···+tnf(xn).
Observation 1.5.3. Iff:[a,b]→Ris a continuous2function on [a,b]and satisﬁes
hypothesis (2)of the proposition, then fis convex.
We have seen that if fsatisﬁes (2), then
f(qx+( 1−q)y)≤qf(x)+( 1 −q)f(y)
for any x,y∈[a,b]a n d q∈[0,1] rational number. Since any real number tcan
be approximated by a sequence of rational numbers qn,a n di ft h e s eq nbelong to
[0,1], we can deduce that
f(qnx+( 1−qn)y)≤qnf(x)+( 1 −qn)f(y).
Now, by using the continuity of fand taking the limit, we get
f(tx+( 1−t)y)≤tf(x)+( 1 −t)f(y).
We say that a function f:[a,b]→Risconcave if−fis convex.
2A function f:[a, b]→Ris continuous at a point c∈[a, b] if lim
x→cf(x)=f(c), and fis
continuous on [ a, b] if it is continous in every point of the interval. Equivalently, fis continuous
atcif for every sequence of points {cn}that converges to c, the sequence {f(cn)}converges to
f(c).

24 Numerical Inequalities
Observation 1.5.4. Af u n c t i o nf :[a,b]→Ris concave if and only if
f(ty+( 1−t)x)≥tf(y)+( 1 −t)f(x)for0≤t≤1anda≤x<y ≤b.
Now, we will consider some criteria to decide whether a function is convex.
Criterion 1.5.5. Af u n c t i o nf :[a,b]→Ris convex if and only if the set {(x, y)|
a≤x≤b,f(x)≤y}is convex.3
Proof. Suppose that fis convex and let A=(x1,y1)a n d B=(x2,y2)b et w o
points in the set U={(x, y)|a≤x≤b,f(x)≤y}.T op r o v et h a t tB+( 1−t)A=
(tx2+( 1−t)x1,t y 2+( 1−t)y1) belongs to U, it is suﬃcient to demonstrate that
a≤tx2+(1−t)x1≤bandf(tx2+(1−t)x1)≤ty2+(1−t)y1. The ﬁrst condition
follows immediately since x1andx2belong to [ a,b].
As for the second condition, since fis convex, it follows that
f(tx2+( 1−t)x1)≤tf(x2)+( 1 −t)f(x1).
Moreover, since f(x2)≤y2andf(x1)≤y1, we can deduce that
f(tx2+( 1−t)x1)≤ty2+( 1−t)y1.
Conversely, we will observe that fis convex if Uis convex.
Letx1,x2∈[a,b] and let us consider A=(x1,f(x1)) and B=(x2,f(x2)).
Clearly AandBbelong to U, and since Uis convex, the segment that joins them
belongs to U, that is, the points of the form tB+( 1− t)Afort∈[0,1]. Thus,
(tx2+( 1−t)x1,t f(x2)+( 1 −t)f(x1))∈U,
but this implies that f(tx2+( 1−t)x1)≤tf(x2)+( 1 −t)f(x1). Hence fis
convex. /square
Criterion 1.5.6. Af u n c t i o n f:[a,b]→Ris convex if and only if, for each x0∈
[a,b], the function P(x)=f(x)−f(x0)
x−x0is non-decreasing for x/negationslash=x0.
Proof. Suppose that fis convex. To prove that P(x) is non-decreasing, we take
x<y and then we show that P(x)≤P(y). One of the following three situations
can arise: x0<x<y ,x<x 0<yorx<y<x 0. Let us consider the ﬁrst of these
3A subset Cof the plane is convex if for any pair of points A,BinC, the segment determined
by these points belongs entirely to C. Since the segment between AandBis the set of points
of the form tB+( 1−t)A,w i t h0 ≤t≤1, the condition is that any point described by this
expression belongs to C.

1.5 Convex functions 25
cases and then the other two can be proved in a similar way. First note that
P(x)≤P(y)⇔f(x)−f(x0)
x−x0≤f(y)−f(x0)
y−x0
⇔(f(x)−f(x0))(y−x0)≤(f(y)−f(x0))(x−x0)
⇔f(x)(y−x0)≤f(y)(x−x0)+f(x0)(y−x)
⇔f(x)≤f(y)x−x0
y−x0+f(x0)y−x
y−x0
⇔f/parenleftbiggx−x0
y−x0y+y−x
y−x0x0/parenrightbigg
≤f(y)x−x0
y−x0+f(x0)y−x
y−x0.
The result follows immediately. /square
Criterion 1.5.7. If the function f:[a,b]→Ris diﬀerentiable4w i t han o n –
decreasing derivative, then fis convex. In particular, if fis twice diﬀerentiable
andf/prime/prime(x)≥0, then the function is convex.
Proof. It is clear that f/prime/prime(x)≥0, forx∈[a,b], implies that f/prime(x) is non-decreasing.
We see that if f/prime(x) is non-decreasing, the function is convex.
Letx=tb+(1−t)abe a point on [ a,b]. Recalling the mean value theorem,5
we know there exist c∈(a,x)a n dd ∈(x, b) such that
f(x)−f(a)=(x−a)f/prime(c)=t(b−a)f/prime(c),
f(b)−f(x)=(b−x)f/prime(d)=( 1 −t)(b−a)f/prime(d).
Then, since f/prime(x) is non-decreasing, we can deduce that
(1−t)(f(x)−f(a)) =t(1−t)(b−a)f/prime(c)≤t(1−t)(b−a)f/prime(d)=t(f(b)−f(x)).
After rearranging terms we get
f(x)≤tf(b)+( 1 −t)f(a). /square
Let us present one geometric interpretation of convexity (and concavity).
Letx,y,zbe points in the interval [ a,b]w i t h x<y<z . If the vertices of
the triangle XYZ have coordinates X=(x, f(x)),Y=(y,f(y)),Z=(z,f(z)),
then the area of the triangle is given by
Δ=1
2detA,w h e r e A=⎛
⎝1xf(x)
1yf(y)
1zf(z)⎞⎠.
4A function f:[a, b]→Ris diﬀerentiable in a point c∈[a, b] if the function f/prime(c)=
lim
x→cf(x)−f(c)
x−cexists and fis diﬀerentiable in A⊂[a, b] if it is diﬀerentiable in every point of A.
5Mean value theorem. For a continuous function f:[a, b]→R, which is diﬀerentiable in ( a, b),
there exists a number x∈(a, b) such that f/prime(x)(b−a)=f(b)−f(a). See [21, page 169].

26 Numerical Inequalities
The area can be positive or negative, this will depend on whether the triangle
XYZ is positively oriented (anticlockwise oriented) or negatively oriented. For a
convex function, we have that Δ >0 and for a concave function, Δ <0, as shown
in the following graphs.
x y z(x, f (x))
(y,f (y))(z,f (z))
x y z(x, f (x))(y,f (y))
(z,f (z))
In fact,
Δ>0⇔detA>0
⇔(z−y)f(x)−(z−x)f(y)+(y−x)f(z)>0
⇔f(y)<z−y
z−xf(x)+y−x
z−xf(z).
If we take t=y−x
z−x,w eh a v e0 <t<1, 1 −t=z−y
z−x,y=tz+( 1− t)xand
f(tz+( 1−t)x)<t f(z)+( 1 −t)f(x).
Now, let us introduce several examples where convex functions are used to
establish inequalities.
Example 1.5.8. The function f(x)=xn,n≥1, is convex in R+and the function
f(x)=xn,w i t h neven, is also convex in R.
This follows from the fact that f/prime/prime(x)=n(n−1)xn−2≥0 in each case.
As an application of this we get the following.
(i) Since/parenleftbiga+b
2/parenrightbig2≤a2+b2
2, we can deduce thata+b
2≤/radicalBig
a2+b2
2,w h i c hi st h e
inequality between the arithmetic mean and the quadratic mean.
(ii) Since/parenleftbiga+b
2/parenrightbign≤an+bn
2, we can deduce that an+bn≥1
2n−1,f o raandb
positive numbers such that a+b=1 .
(iii) If aandbare positive numbers,/parenleftbig
1+a
b/parenrightbign+/parenleftbig
1+b
a/parenrightbign≥2n+1. This follows

1.5 Convex functions 27
from
2n=f(2)≤f/parenleftBigg
a+b
a+a+b
b
2/parenrightBigg
≤1
2/bracketleftbigg
f/parenleftBig
1+a
b/parenrightBig
+f/parenleftbigg
1+b
a/parenrightbigg/bracketrightbigg
=1
2/bracketleftbigg/parenleftBig
1+a
b/parenrightBign
+/parenleftbigg
1+b
a/parenrightbiggn/bracketrightbigg
.
Example 1.5.9. The exponential function f(x)=exis convex in R,s i n c e f/prime/prime(x)=
ex>0, for every x∈R.
Let us observe several ways in which this property can be used.
(i)(Weighted AM-GM inequality) Ifx1,…,xn,t1,…,tnare positive numbers
and/summationtextn
i=1ti=1 ,t h e n
xt1
1···xtn
n≤t1x1+···+tnxn.
In fact, since xti
i=etilogxiandexis convex, we can deduce that
xt1
1···xtn
n=et1logx1···etnlogxn=et1logx1+···+tnlogxn
≤t1elogx1+···+tnelogxn=t1x1+···+tnxn.
In particular, if we take ti=1
n,f o r1 ≤i≤n, we can produce another proof
of the inequality between the arithmetic mean and the geometric mean for n
numbers.
(ii)(Young’s inequality) Letx,ybe positive real numbers. If a,b>0s a t i s f yt h e
condition1
a+1
b=1 ,t h e n xy≤1
axa+1
byb.
We only need to apply part (i) as follows:
xy=(xa)1
a/parenleftbig
yb/parenrightbig1
b≤1
axa+1
byb.
(iii)(H¨older’s inequality) Letx1,x2,…,xn,y1,y2,…,ynbe positive numbers
anda,b>0 such that1
a+1
b=1 ,t h e n
n/summationdisplay
i=1xiyi≤/parenleftBiggn/summationdisplay
i=1xa
i/parenrightBigg1/a/parenleftBiggn/summationdisplay
i=1yb
i/parenrightBigg1/b
.
Let us ﬁrst assume that/summationtextn
i=1xa
i=/summationtextn
i=1yb
i=1 .
Using part (ii), xiyi≤1
axa
i+1
byb
i,t h e n
n/summationdisplay
i=1xiyi≤1
an/summationdisplay
i=1xa
i+1
bn/summationdisplay
i=1yb
i=1
a+1
b=1.

28 Numerical Inequalities
Now, suppose that/summationtextn
i=1xa
i=Aand/summationtextn
i=1yb
i=B.L e tu st a k e x/prime
i=xi
A1/a
andy/prime
i=yi
B1/b.S i n c e
n/summationdisplay
i=1(x/prime
i)a=/summationtextn
i=1xa
i
A=1 a n dn/summationdisplay
i=1(y/prime
i)b=/summationtextn
i=1yb
i
B=1 ,
we can deduce that
1≥n/summationdisplay
i=1x/prime
iy/prime
i=n/summationdisplay
i=1xiyi
A1/aB1/b=1
A1/aB1/bn/summationdisplay
i=1xiyi.
Therefore,/summationtextn
i=1xiyi≤A1/aB1/b.
If we choose a=b= 2, we get the Cauchy-Schwarz inequality.
Let us introduce a consequence of H¨ older’s inequality, which is a generaliza-
tion of the triangle inequality.
Example 1.5.10 (Minkowski’s inequality). Leta1,a2,…,an,b1,b2,…,bnbe
positive numbers and p>1,t h e n
/parenleftBiggn/summationdisplay
k=1(ak+bk)p/parenrightBigg1
p
≤/parenleftBiggn/summationdisplay
k=1(ak)p/parenrightBigg1
p
+/parenleftBiggn/summationdisplay
k=1(bk)p/parenrightBigg1
p
.
We note that
(ak+bk)p=ak(ak+bk)p−1+bk(ak+bk)p−1,
so that
n/summationdisplay
k=1(ak+bk)p=n/summationdisplay
k=1ak(ak+bk)p−1+n/summationdisplay
k=1bk(ak+bk)p−1. (1.6)
We apply H¨ older’s inequality to each term of the sum on the right-hand side of
(1.6), with qsuch that1
p+1
q=1 ,t og e t
n/summationdisplay
k=1ak(ak+bk)p−1≤/parenleftBiggn/summationdisplay
k=1(ak)p/parenrightBigg1
p/parenleftBiggn/summationdisplay
k=1(ak+bk)q(p−1)/parenrightBigg1
q
,
n/summationdisplay
k=1bk(ak+bk)p−1≤/parenleftBiggn/summationdisplay
k=1(bk)p/parenrightBigg1
p/parenleftBiggn/summationdisplay
k=1(ak+bk)q(p−1)/parenrightBigg1
q
.
Putting these inequalities into (1.6), and noting that q(p−1) = p, yields the
required inequality. Note that Minkowski’s inequality is an equality if we allow
p=1 .F o r0 <p< 1, the inequality is reversed.

1.5 Convex functions 29
Example 1.5.11. (Short list IMO, 1998) Ifr1,…,rnare real numbers greater than
1, prove that
1
1+r1+···+1
1+rn≥n
n√
r1···rn+1.
First note that the function f(x)=1
1+exis convex for R+,s i n c e f/prime(x)=
−ex
(1+ex)2andf/prime/prime(x)=ex(ex−1)
(ex+1)3≥0f o rx> 0.
Now, if ri>1, then ri=exifor some xi>0. Since f(x)=1
1+exis convex,
w ec a ne s t a b l i s ht h a t
1
e(x1+···+xn
n)+1≤1
n/parenleftbigg1
1+ex1+···+1
1+exn/parenrightbigg
,
hence
n
n√
r1···rn+1≤1
1+r1+···+1
1+rn.
Example 1.5.12. (China, 1989) Prove that for any nreal positive numbers x1,…,
xnsuch that/summationtextn
i=1xi=1, we have
n/summationdisplay
i=1xi
√
1−xi≥/summationtextni=1√
xi
√
n−1.
We will use the fact that the function f(x)=x
√
1−xis convex in (0 ,1), since
f/prime/prime(x)>0,
1
nn/summationdisplay
i=1xi
√
1−xi=1
nn/summationdisplay
i=1f(xi)≥f/parenleftBiggn/summationdisplay
i=11
nxi/parenrightBigg
=f/parenleftbigg1
n/parenrightbigg
=1
√
n√
n−1,
hence
n/summationdisplay
i=1xi
√
1−xi≥√
n
√
n−1.
It is left to prove that/summationtextn
i=1√
xi≤√
n, but this follows from the Cauchy-Schwarz
inequality,/summationtextn
i=1√
xi≤/radicalbig
/summationtextni=1xi/radicalbig
/summationtextni=11=√
n.
Example 1.5.13. (Hungary–Israel, 1999) Letkandlbe two given positive integers,
and let aij,1≤i≤kand1≤j≤l,b eklgiven positive numbers. Prove that if
q≥p>0,t h e n
⎛
⎝l/summationdisplay
j=1/parenleftBiggk/summationdisplay
i=1ap
ij/parenrightBiggq
p⎞
⎠1
q
≤⎛
⎜⎝k/summationdisplay
i=1⎛
⎝l/summationdisplay
j=1aq
ij⎞
⎠p
q⎞
⎟⎠1
p
.

30 Numerical Inequalities
Deﬁne bj=/summationtextk
i=1ap
ijforj=1,2,…,l , and denote the left-hand side of the
required inequality by Land the right-hand side by R.T h e n
Lq=l/summationdisplay
j=1bq
p
j
=l/summationdisplay
j=1/parenleftBigg
bq−p
p
j/parenleftBiggk/summationdisplay
i=1apij/parenrightBigg/parenrightBigg
=k/summationdisplay
i=1⎛
⎝l/summationdisplay
j=1bq−p
p
jap
ij⎞
⎠.
Using H¨ older’s inequality we obtain
Lq≤k/summationdisplay
i=1⎡
⎢⎣⎛
⎝l/summationdisplay
j=1/parenleftbigg
bq−p
p
j/parenrightbigg q
q−p⎞⎠q−p
q⎛⎝
l/summationdisplay
j=1(ap
ij)q
p⎞
⎠p
q⎤
⎥⎦
=k/summationdisplay
i=1⎡
⎢⎣⎛
⎝l/summationdisplay
j=1bq
p
j⎞⎠q−p
q⎛⎝
l/summationdisplay
j=1aq
ij⎞
⎠p
q⎤
⎥⎦
=⎛
⎝l/summationdisplay
j=1bq
p
j⎞⎠q−p
q
·⎡
⎢⎣k/summationdisplay
i=1⎛
⎝l/summationdisplay
j=1aq
ij⎞
⎠p
q⎤
⎥⎦=Lq−pRp.
The inequality L≤Rfollows by dividing both sides of Lq≤Lq−pRpbyLq−pand
taking the p-th root.
Exercise 1.77. (i) For a,b∈R+,w i t h a+b=1 ,p r o v et h a t
/parenleftbigg
a+1
a/parenrightbigg2
+/parenleftbigg
b+1
b/parenrightbigg2
≥25
2.
(ii) For a,b,c∈R+,w i t h a+b+c=1 ,p r o v et h a t
/parenleftbigg
a+1
a/parenrightbigg2
+/parenleftbigg
b+1
b/parenrightbigg2
+/parenleftbigg
c+1
c/parenrightbigg2
≥100
3.
Exercise 1.78. For 0≤a,b,c≤1, prove that
a
b+c+1+b
c+a+1+c
a+b+1+( 1−a)(1−b)(1−c)≤1.

1.5 Convex functions 31
Exercise 1.79. (Russia, 2000) For real numbers x,ysuch that 0 ≤x,y≤1, prove
that1
√
1+x2+1
/radicalbig
1+y2≤2
√
1+xy.
Exercise 1.80. Prove that the function f(x)=s i n xis concave in the interval [0 ,π].
Use this to verify that the angles A, B, C of a triangle satisfy sin A+sinB+sinC≤
3
2√
3.
Exercise 1.81. IfA,B,C,Dare angles belonging to the interval [0 ,π], then
(i) sin AsinB≤sin2/parenleftbigA+B
2/parenrightbig
and the equality holds if and only if A=B,
(ii) sin AsinBsinCsinD≤sin4/parenleftbigA+B+C+D
4/parenrightbig
,
(iii) sin AsinBsinC≤sin3/parenleftbigA+B+C
3/parenrightbig
,
Moreover, if A, B, C are the internal angles of a triangle, then
(iv) sin AsinBsinC≤3
8√
3,
(v) sinA
2sinB
2sinC
2≤1
8,
(vi) sin A+s i nB+s i nC=4c o sA
2cosB
2cosC
2.
Exercise 1.82. (Bernoulli’s inequality)
(i) For any real number x>−1 and for every positive integer n,w eh a v e( 1+
x)n≥1+nx.
(ii) Use this inequality to provide another proof of the AM-GM inequality.
Exercise 1.83. (Sch¨ ur’s inequality) Ifx,y,zare positive real numbers and nis a
positive integer, we have
xn(x−y)(x−z)+yn(y−z)(y−x)+zn(z−x)(z−y)≥0.
For the case n= 1, the inequality can take one of the following forms:
(a)x3+y3+z3+3xyz≥xy(x+y)+yz(y+z)+zx(z+x).
(b)xyz≥(x+y−z)(y+z−x)(z+x−y).
(c) If x+y+z=1 ,9 xyz+1≥4(xy+yz+zx).
Exercise 1.84. (Canada, 1992) For any three non-negative real numbers x,yand
zwe have
x(x−z)2+y(y−z)2≥(x−z)(y−z)(x+y−z).
Exercise 1.85. Ifa,b,care positive real numbers, prove that
a
(b+c)2+b
(c+a)2+c
(a+b)2≥9
4(a+b+c).

32 Numerical Inequalities
Exercise 1.86. Leta,bandcbe positive real numbers, prove that
1+3
ab+bc+ca≥6
a+b+c.
Moreover, if abc=1 ,p r o v et h a t
1+3
a+b+c≥6
ab+bc+ca.
Exercise 1.87. (Power mean inequality) Letx1,x2,…,xnbe positive real numbers
and let t1,t2,…,tnbe positive real numbers adding up to 1. Let randsbe two
nonzero real numbers such that r>s.P r o v et h a t
(t1xr
1+···+tnxrn)1
r≥(t1xs1+···+tnxsn)1
s
with equality if and only if x1=x2=···=xn.
Exercise 1.88. (Two extensions of H¨ older’s inequality) Letx1,x2,…,xn,y1,y2,
…,yn,z1,z2,…,znbe positive real numbers.
(i) If a,b,care positive real numbers such that1
a+1
b=1
c,t h e n
/braceleftBiggn/summationdisplay
i=1(xiyi)c/bracerightBigg1
c
≤/braceleftBiggn/summationdisplay
i=1xia/bracerightBigg1
a/braceleftBiggn/summationdisplay
i=1yib/bracerightBigg1
b
.
(ii) If a,b,care positive real numbers such that1
a+1
b+1
c=1 ,t h e n
n/summationdisplay
i=1xiyizi≤/braceleftBiggn/summationdisplay
i=1xia/bracerightBigg1
a/braceleftBiggn/summationdisplay
i=1yib/bracerightBigg1
b/braceleftBiggn/summationdisplay
i=1zic/bracerightBigg1
c
.
Exercise 1.89. (Popoviciu’s inequality) IfIis an interval and f:I→Ris a convex
function, then for a,b,c∈Ithe following inequality holds:
2
3/bracketleftbigg
f/parenleftbigga+b
2/parenrightbigg
+f/parenleftbiggb+c
2/parenrightbigg
+f/parenleftbiggc+a
2/parenrightbigg/bracketrightbigg
≤f(a)+f(b)+f(c)
3+f/parenleftbigga+b+c
3/parenrightbigg
.
Exercise 1.90. Leta,b,cbe non-negative real numbers. Prove that
(i)a2+b2+c2+33√
a2b2c2≥2(ab+bc+ca),
(ii)a2+b2+c2+2abc+1≥2(ab+bc+ca).
Exercise 1.91. Leta,b,cbe positive real numbers. Prove that
/parenleftbiggb+c
a+c+a
b+a+b
c/parenrightbigg
≥4/parenleftbigga
b+c+b
c+a+c
a+b/parenrightbigg
.

1.6 A helpful inequality 33
1.6 A helpful inequality
First, let us study two very useful algebraic identities that are deduced by consid-
ering a special factor of a3+b3+c3−3abc.
LetPdenote the cubic polynomial
P(x)=x3−(a+b+c)x2+(ab+bc+ca)x−abc,
which has a,bandcas its roots. By substituting a,b,cin the polynomial, we
obtain
a3−(a+b+c)a2+(ab+bc+ca)a−abc=0 ,
b3−(a+b+c)b2+(ab+bc+ca)b−abc=0 ,
c3−(a+b+c)c2+(ab+bc+ca)c−abc=0.
Adding up these three equations yields
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca). (1.7)
It immediately follows that if a+b+c=0 ,t h e n a3+b3+c3=3abc.
Note also that the expression
a2+b2+c2−ab−bc−ca
can also be written as
a2+b2+c2−ab−bc−ca=1
2[(a−b)2+(b−c)2+(c−a)2].(1.8)
In this way, we obtain another version of identity (1.7),
a3+b3+c3−3abc=1
2(a+b+c)[(a−b)2+(b−c)2+(c−a)2]. (1.9)
This presentation of the identity leads to a short proof of the AM-GM inequalityfor three variables. From (1.9) it is clear that if a,b,care positive numbers, then
a
3+b3+c3≥3abc.N o w ,i f x,y,zare positive numbers, taking a=3√
x,b=3√
y
andc=3√
zwill lead us to
x+y+z
3≥3√
xyz
with equality if and only if x=y=z.
Note that identity (1.8) provides another proof of Exercise 1.27.
Exercise 1.92. For real numbers x,y,z,p r o v et h a t
x2+y2+z2≥|xy+yz+zx|.

34 Numerical Inequalities
Exercise 1.93. For positive real numbers a,b,c,p r o v et h a t
a2+b2+c2
abc≥1
a+1
b+1
c.
Exercise 1.94. Ifx,y,zare real numbers such that x<y<z ,p r o v et h a t
(x−y)3+(y−z)3+(z−x)3>0.
Exercise 1.95. Leta,b,cbe the side lengths of a triangle. Prove that
3/radicalbigg
a3+b3+c3+3abc
2≥max{a,b,c}.
Exercise 1.96. (Romania, 2007) For non-negative real numbers x,y,z,p r o v et h a t
x3+y3+z3
3≥xyz+3
4|(x−y)(y−z)(z−x)|.
Exercise 1.97. (UK, 2008) Find the minimum of x2+y2+z2,w h e r e x,y,zare
real numbers such that x3+y3+z3−3xyz=1 .
A very simple inequality which may be helpful for proving a large number of
algebraic inequalities is the following.
Theorem 1.6.1 (A helpful inequality). Ifa,b,x,yare real numbers and x,y>0,
then the following inequality holds:
a2
x+b2
y≥(a+b)2
x+y. (1.10)
Proof. The proof is quite simple. Clearing out denominators, we can express the
inequality as
a2y(x+y)+b2x(x+y)≥(a+b)2xy,
which simpliﬁes to become the obvious ( ay−bx)2≥0. We see that the equality
holds if and only if ay=bx, that is, if and only ifa
x=b
y.
Another form to prove the inequality is using the Cauchy-Schwarz inequality
in the following way:
(a+b)2=/parenleftbigga
√
x√
x+b
√
y√
y/parenrightbigg2
≤/parenleftbigga2
x+b2
y/parenrightbigg
(x+y). /square
Using the above theorem twice, we can extend the inequality to three pairs
of numbers
a2
x+b2
y+c2
z≥(a+b)2
x+y+c2
z≥(a+b+c)2
x+y+z,

1.6 A helpful inequality 35
and a simple inductive argument shows that
a2
1
x1+a22
x2+···+a2n
xn≥(a1+a2+···+an)2
x1+x2+···+xn(1.11)
for all real numbers a1,a2,…,anandx1,x2,…,xn>0, with equality if and
only ifa1
x1=a2
x2=···=an
xn.
Inequality (1.11) is also called the Cauchy-Schwarz inequality in Engel form or
Arthur Engel’s Minima Principle.
As a ﬁrst application of this inequality, we will present another proof of the
Cauchy-Schwarz inequality. Let us write
a2
1+a22+···+a2n=a2
1b21
b21+a22b22
b22+···+a2nb2n
b2n,
then
a2
1b21
b2
1+a2
2b22
b2
2+···+a2
nb2n
b2n≥(a1b1+a2b2+···+anbn)2
b2
1+b22+···+b2n.
Thus, we conclude that
(a2
1+a22+···+a2n)(b21+b22+···+b2n)≥(a1b1+a2b2+···+anbn)2
and the equality holds if and only if
a1
b1=a2
b2=···=an
bn.
It is worth to mention that there are other forms of the Cauchy-Schwarz
inequality in Engel form.
Example 1.6.2. Leta1,…,an,b1,…,bnbe positive real numbers. Prove that
(i)a1
b1+···+an
bn≥(a1+···+an)2
a1b1+···+anbn,
(ii)a1
b2
1+···+an
b2n≥1
a1+···+an/parenleftbigga1
b1+···+an
bn/parenrightbigg2
.
Both inequalities are direct consequence of inequality (1.11), as we can see
as follows.
(i)a1
b1+···+an
bn=a2
1
a1b1+···+a2n
anbn≥(a1+···+an)2
a1b1+···+anbn,
(ii)a1
b21+···+an
b2n=a2
1
b2
1
a1+···+a2
n
b2n
an≥1
a1+···+an/parenleftbigga1
b1+···+an
bn/parenrightbigg2
.

36 Numerical Inequalities
Example 1.6.3. (APMO, 1991) Leta1,…,an,b1,…,bnbe positive real numbers
such that a1+a2+···+an=b1+b2+···+bn. Prove that
a2
1
a1+b1+···+a2
n
an+bn≥1
2(a1+···+an).
Observe that (1.11) implies that
a2
1
a1+b1+···+a2n
an+bn≥(a1+a2+···+an)2
a1+a2+···+an+b1+b2+···+bn
=(a1+a2+···+an)2
2(a1+a2+···+an)
=1
2(a1+a2+···+an).
The following example consists of a proof of the quadratic mean-arithmetic
mean inequality.
Example 1.6.4 (Quadratic mean-arithmetic mean inequality). For positive real
numbers x1,…,xn, we have
/radicalbigg
x2
1+x22+···+x2n
n≥x1+x2+···+xn
n.
Observe that using (1.11) leads us to
x2
1+x22+···+x2n
n≥(x1+x2+···+xn)2
n2,
which implies the above inequality.
In some cases the numerators are not squares, but a simple trick allows us
to write them as squares, so that we can use the inequality. Our next application
shows this trick and oﬀers a shorter proof for Example 1.4.9.
Example 1.6.5. (IMO, 1995) Leta,b,cbe positive real numbers such that abc=1.
Prove that1
a3(b+c)+1
b3(a+c)+1
c3(a+b)≥3
2.
Observe that
1
a3(b+c)+1
b3(c+a)+1
c3(a+b)=1
a2
a(b+c)+1
b2
b(c+a)+1
c2
c(a+b)
≥(1
a+1
b+1
c)2
2(ab+bc+ca)=ab+bc+ca
2(abc)
≥33/radicalbig
(abc)2
2=3
2,

1.6 A helpful inequality 37
where the ﬁrst inequality follows from (1.11) and the second is a consequence of
theAM-GMinequality.
As a further example of the use of inequality (1.11), we provide a simple
proof of Nesbitt’s inequality.
Example 1.6.6 (Nesbitt’s inequality). Fora,b,c∈R+, we have
a
b+c+b
c+a+c
a+b≥3
2.
We multiply the three terms on the left-hand side of the inequality bya
a,b
b,
c
c, respectively, and then we use inequality (1.11) to produce
a2
a(b+c)+b2
b(c+a)+c2
c(a+b)≥(a+b+c)2
2(ab+bc+ca).
From Equation (1.8) we know that a2+b2+c2−ab−bc−ca≥0, that is,
(a+b+c)2≥3(ab+bc+ca). Therefore
a
b+c+b
c+a+c
a+b≥(a+b+c)2
2(ab+bc+ca)≥3
2.
Example 1.6.7. (Czech and Slovak Republics, 1999) Fora,bandcpositive real
numbers, prove the inequality
a
b+2c+b
c+2a+c
a+2b≥1.
Observe that
a
b+2c+b
c+2a+c
a+2b=a2
ab+2ca+b2
bc+2ab+c2
ca+2bc.
Then using (1.11) yields
a2
ab+2ca+b2
bc+2ab+c2
ca+2bc≥(a+b+c)2
3(ab+bc+ca)≥1,
where the last inequality follows in the same way as in the previous example.Exercise 1.98. (South Africa, 1995) For a,b,c,dpositive real numbers, prove that
1
a+1
b+4
c+16
d≥64
a+b+c+d.
Exercise 1.99. Letaandbbe positive real numbers. Prove that
8(a4+b4)≥(a+b)4.

38 Numerical Inequalities
Exercise 1.100. Letx,y,zbe positive real numbers. Prove that
2
x+y+2
y+z+2
z+x≥9
x+y+z.
Exercise 1.101. Leta,b,x,y,zbe positive real numbers. Prove that
x
ay+bz+y
az+bx+z
ax+by≥3
a+b.
Exercise 1.102. Leta,b,cbe positive real numbers. Prove that
a2+b2
a+b+b2+c2
b+c+c2+a2
c+a≥a+b+c.
Exercise 1.103. (i) Let x,y,zbe positive real numbers. Prove that
x
x+2y+3z+y
y+2z+3x+z
z+2x+3y≥1
2.
(ii) (Moldova, 2007) Let w,x,y,zbe positive real numbers. Prove that
w
x+2y+3z+x
y+2z+3w+y
z+2w+3x+z
w+2x+3y≥2
3.
Exercise 1.104. (Croatia, 2004) Let x,y,zbe positive real numbers. Prove that
x2
(x+y)(x+z)+y2
(y+z)(y+x)+z2
(z+x)(z+y)≥3
4.
Exercise 1.105. Fora,b,c,dpositive real numbers, prove that
a
b+c+b
c+d+c
d+a+d
a+b≥2.
Exercise 1.106. Leta,b,c,d,ebe positive real numbers. Prove that
a
b+c+b
c+d+c
d+e+d
e+a+e
a+b≥5
2.
Exercise 1.107. (i) Prove that, for all positive real numbers a,b,c,x,y,zwith
a≥b≥candz≥y≥x, the following inequality holds:
a3
x+b3
y+c3
z≥(a+b+c)3
3(x+y+z).
(ii) (Belarus, 2000) Prove that, for all positive real numbers a,b,c,x,y,z,t h e
following inequality holds:
a3
x+b3
y+c3
z≥(a+b+c)3
3(x+y+z).

1.7 The substitution strategy 39
Exercise 1.108. (Greece, 2008) For x1,x2,…,xnpositive integers, prove that
/parenleftbiggx2
1+x22+···+x2n
x1+x2+···+xn/parenrightbiggkn
t
≥x1·x2·····xn,
where k=m a x {x1,x2,…,x n}andt=m i n {x1,x2,…,x n}. Under which con-
dition the equality holds?
1.7 The substitution strategy
Substitution is a useful strategy to solve inequality problems. Making an adequate
substitution we can, for instance, change the diﬃcult terms of the inequality a
little, we can simplify expressions or we can reduce terms. In this section we give
some ideas of what can be done with this strategy. As always, the best way to dothat is through some examples.
One useful suggestion for problems that contain in the hypothesis an extra
condition, is to use that condition to simplify the problem. In the next example we
apply this technique to eliminate the denominators in order to make the problem
easier to solve.
Example 1.7.1. Ifa,b,care positive real numbers less than 1,w i t h a+b+c=2,
then /parenleftbigga
1−a/parenrightbigg/parenleftbiggb
1−b/parenrightbigg/parenleftbiggc
1−c/parenrightbigg
≥8.
After performing the substitution x=1−a,y=1−b,z=1−c,w eo b t a i n
thatx+y+z=3−(a+b+c)=1 , a=1−x=y+z,b=z+x,c=x+y. Hence
the inequality is equivalent to
/parenleftbiggy+z
x/parenrightbigg/parenleftbiggz+x
y/parenrightbigg/parenleftbiggx+y
z/parenrightbigg
≥8,
and in turn, this is equivalent to
(x+y)(y+z)(z+x)≥8xyz.
This last inequality is quite easy to prove. It is enough to apply three times the
AM-GM inequality under the form ( x+y)≥2√
xy(see Exercise 1 .26).
It may be possible that the extra condition is used only as part of the solution,
as in the following two examples.
Example 1.7.2. (Mexico, 2007) Ifa,b,care positive real numbers that satisfy
a+b+c=1, prove that
√
a+bc+√
b+ca+√
c+ab≤2.

40 Numerical Inequalities
Using the condition a+b+c=1 ,w eh a v et h a t
a+bc=a(a+b+c)+bc=(a+b)(a+c),
then, by the AM-GM inequality it follows that
√
a+bc=/radicalbig
(a+b)(a+c)≤2a+b+c
2.
Similarly,
√
b+ca≤2b+c+a
2and√
c+ab≤2c+a+b
2.
Thus, after adding the three inequalities we obtain
√
a+bc+√
b+ca+√
c+ab
≤2a+b+c
2+2b+c+a
2+2c+a+b
2=4a+4b+4c
2=2.
The equality holds when a+b=a+c,b+c=b+aandc+a=c+b,t h a ti s ,
when a=b=c=1
3.
Example 1.7.3. Ifa,b,care positive real numbers with ab+bc+ca=1,p r o v e
thata
√
a2+1+b
√
b2+1+c
√
c2+1≤3
2.
Note that ( a2+1 )= a2+ab+bc+ca=(a+b)(a+c). Similarly, b2+1=
(b+c)(b+a)a n d c2+1=( c+a)(c+b). Now, the inequality under consideration
is equivalent to
a
/radicalbig
(a+b)(a+c)+b
/radicalbig
(b+c)(b+a)+c
/radicalbig
(c+a)(c+b)≤3
2.
Using the AM-GM inequality, applied to every element of the sum on the left-hand
side, we obtain
a
/radicalbig
(a+b)(a+c)+b
/radicalbig
(b+c)(b+a)+c
/radicalbig
(c+a)(c+b)
≤1
2/parenleftbigga
a+b+a
a+c/parenrightbigg
+1
2/parenleftbiggb
b+c+b
b+a/parenrightbigg
+1
2/parenleftbiggc
c+a+c
c+b/parenrightbigg
=3
2.
Many inequality problems suggest which substitution should be made. In the
following example the substitution allows us to make at least one of the terms inthe inequality look simpler.
Example 1.7.4. (India, 2002) Ifa,b,care positive real numbers, prove that
a
b+b
c+c
a≥c+a
c+b+a+b
a+c+b+c
b+a.

1.7 The substitution strategy 41
Making the substitution x=a
b,y=b
c,z=c
athe left-hand side of the
inequality is now more simple, x+y+z.L e tu ss e eh o wt h er i g h t – h a n ds i d e
changes. The ﬁrst element of the sum is modiﬁed as follows:
c+a
c+b=1+a
c
1+b
c=1+a
bb
c
1+b
c=1+xy
1+y=x+1−x
1+y.
Similarly,
a+b
a+c=y+1−y
1+zandb+c
b+a=z+1−z
1+x.
Now, the inequality is equivalent to
x−1
1+y+y−1
1+z+z−1
1+x≥0
with the extra condition xyz=1 .
The last inequality can be rewritten as
(x2−1)(z+1 )+( y2−1)(x+1 )+( z2−1)(y+1 )≥0,
w h i c hi nt u r ni se q u i v a l e n tt o
x2z+y2x+z2y+x2+y2+z2≥x+y+z+3.
But, from the AM-GM inequality, we have x2z+y2x+z2y≥33/radicalbig
x3y3z3=3 .A l s o ,
x2+y2+z2≥1
3(x+y+z)2=x+y+z
3(x+y+z)≥3√
xyz(x+y+z)=x+y+z,
where the ﬁrst inequality follows from inequality (1.11).
In order to make a substitution, sometimes it is necessary to work a little bit
beforehand, as we can see in the following example. This example also helps us
to point out that we may need to make more than one substitution in the same
problem.
Example 1.7.5. Leta,b,cbe positive real numbers, prove that
(a+b)(a+c)≥2/radicalbig
abc(a+b+c).
Dividing both sides of the given inequality by a2and setting x=b
a,y=c
a,
the inequality becomes
(1 +x)(1 + y)≥2/radicalbig
xy(1 +x+y).
Now, dividing both sides by xyand making the substitution r=1+1
x,s=1+1
y,
the inequality we need to prove becomes
rs≥2√
rs−1.
This last inequality is equivalent to ( rs−2)2≥0, which become evident after
squaring both sides and doing some algebra.
It is a common situation for inequality problems to have several solutions
and also to accept several substitutions that help to solve the problem. We will
see an instance of this in the next example.

42 Numerical Inequalities
Example 1.7.6. (Korea, 1998) Ifa,b,care positive real numbers such that a+b+
c=abc, prove that
1
√
1+a2+1
√
1+b2+1
√
1+c2≤3
2.
Under the substitution x=1
a,y=1
b,z=1
c, condition a+b+c=abc
becomes xy+yz+zx= 1 and the inequality becomes equivalent to
x
√
x2+1+y
/radicalbig
y2+1+z
√
z2+1≤3
2.
This is the third example in this section.
Another solution is to make the substitution a=t a n A,b=t a n B,c=t a n C.
Since tan A+t a n B+t a n C=t a n AtanBtanC,t h e n A+B+C=π(or a
multiple of π). Now, since 1 + tan2A=( c o s A)−2, the inequality is equivalent to
cosA+c o s B+c o s C≤3
2, which is a valid result as will be shown in Example
2.5.2. Note that the Jensen inequality cannot be applied in this case because the
function f(x)=1
√
1+x2is not concave in R+.
We note that not all substitutions are algebraic, since there are trigonometric
substitutions that can be useful, as is shown in the last example and as we willsee next. Also, as will be shown in Sections 2 .2a n d2 .5 of the next chapter, there
are some geometric substitutions that can be used for the same purposes.
Example 1.7.7. (Romania, 2002) Ifa,b,care real numbers in the interval (0,1),
prove that√
abc+/radicalbig
(1−a)(1−b)(1−c)<1.
Making the substitution a=c o s2A,b=c o s2B,c=c o s2C,w i t h A,B,Cin
the interval (0 ,π
2), we obtain that√
1−a=√
1−cos2A=s i nA,√
1−b=s i nB
and√
1−c=s i nC. Therefore the inequality is equivalent to
cosAcosBcosC+s i nAsinBsinC<1.
But observe that
cosAcosBcosC+s i nAsinBsinC<cosAcosB+s i nAsinB
=c o s ( A−B)≤1.
Exercise 1.109. Letx,y,zbe positive real numbers. Prove that
x3
x3+2y3+y3
y3+2z3+z3
z3+2×3≥1.
Exercise 1.110. (Kazakhstan, 2008) Let x,y,zbe positive real numbers such that
xyz=1 .P r o v et h a t
1
yz+z+1
zx+x+1
xy+y≥3
2.

1.8 Muirhead’s theorem 43
Exercise 1.111. (Russia, 2004) If n>3a n d x1,x2,…,x nare positive real numbers
withx1x2···xn=1 ,p r o v et h a t
1
1+x1+x1x2+1
1+x2+x2x3+···+1
1+xn+xnx1>1.
Exercise 1.112. (Poland, 2006) Let a,b,cbe positive real numbers such that
ab+bc+ca=abc.P r o v et h a t
a4+b4
ab(a3+b3)+b4+c4
bc(b3+c3)+c4+a4
ca(c3+a3)≥1.
Exercise 1.113. (Ireland, 2007) Let a,b,cbe positive real numbers, prove that
1
3/parenleftbiggbc
a+ca
b+ca
b/parenrightbigg
≥/radicalbigg
a2+b2+c2
3≥a+b+c
3.
Exercise 1.114. (Romania, 2008) Let a,b,cbe positive real numbers with abc=8 .
Prove that
a−2
a+1+b−2
b+1+c−2
c+1≤0.
1.8 Muirhead’s theorem
In 1903, R.F. Muirhead published a paper containing the study of some algebraic
methods applicable to identities and inequalities of symmetric algebraic functionsofnvariables.
While considering algebraic expressions of the form x
a1
1xa2
2···xann,h ea n –
alyzed symmetric polynomials containi ng these expressions in order to create a
“certain order” in the space of n-tuples ( a1,a2,…,a n) satisfying the condition
a1≥a2≥···≥ an.
We will assume that xi>0 for all 1 ≤i≤n. We will denote by
/summationdisplay
!F(x1,…,x n)
the sum of the n! terms obtained from evaluating Fin all possible permutations
of (x1,…,x n). We will consider only the particular case
F(x1,…,x n)=xa1
1xa2
2···xan
nwithxi>0,ai≥0.
We write [ a]=[a1,a2,…,a n]=1
n!/summationdisplay
!xa1
1xa2
2···xann. For instance, for the vari-
ables x,y,z>0w eh a v et h a t
[1,1] =xy,[ 1,1,1] =xyz,[ 2,1,0] =1
3![x2(y+z)+y2(x+z)+z2(x+y)].

44 Numerical Inequalities
It is clear that [a ] is invariant under any permutation of the ( a1,a2,…,a n)a n d
therefore two sets of aare the same if they only diﬀer in arrangement. We will
say that a mean value of the type [ a] is a symmetrical mean. In particular,
[1,0,…,0] =(n−1)!
n!(x1+x2+···+xn)=1
n/summationtextn
i=1xiis the arithmetic mean
and [1
n,1
n,…,1
n]=n!
n!(x1
n
1·x1
n
2···x1
nn)=n√
x1x2···xnis the geometric mean.
When a1+a2+···+an=1 ,[a] is a common generalization of both the arithmetic
mean and the geometric mean.
Ifa1≥a2≥···≥ anandb1≥b2≥···≥ bn, usually [ b] is not comparable
to [a], in the sense that there is an inequality between their associated expressions
valid for all n-tuples of non-negative real numbers x1,x2,…,xn.
Muirhead wanted to compare the values of the symmetric polynomials [ a]a n d
[b] for any set of non-negative values of the variables occurring in both polynomials.
From now on we denote ( a)=(a1,a2,…,a n).
Deﬁnition 1.8.1. We will say that (b)≺(a)((b)is majorized by (a))w h e n (a)and
(b)can be rearranged to satisfy the following two conditions:
(1)n/summationdisplay
i=1bi=n/summationdisplay
i=1ai;
(2)ν/summationdisplay
i=1bi≤ν/summationdisplay
i=1aifor all 1≤ν<n.
It is clear that ( a)≺(a)a n dt h a t( b)≺(a)a n d( c)≺(b) implies ( c)≺(a).
Theorem 1.8.2 (Muirhead’s theorem). [b]≤[a]for any n-tuple of non-negative
numbers (x1,x2,…,x n)if and only if (b)≺(a). Equality takes place only when
(b)and(a)are identical or when all the xisa r ee q u a l .
Before going through the proof, which is quite diﬃcult, let us look at some
examples. First, it is clear that [2 ,0,0] cannot be compared with [1 ,1,1] because
the ﬁrst condition in Deﬁnition 1.8.1 is not satisﬁed, but we can see that [2 ,0,0]≥
[1,1,0], which is equivalent to
x2+y2+z2≥xy+yz+zx.
I nt h es a m ew a y ,w ec a ns e et h a t
1.×2+y2≥2xy⇔[2,0]≥[1,1],
2.×3+y3+z3≥3xyz⇔[3,0,0]≥[1,1,1],
3.×5+y5≥x3y2+x2y3⇔[5,0]≥[3,2],
4.x2y2+y2z2+z2x2≥x2yz+y2xz+z2xy⇔[2,2,0]≥[2,1,1],

1.8 Muirhead’s theorem 45
and all these inequalities are satisﬁed if we take for granted Muirhead’s theorem.
Proof of Muirhead’s theorem. Suppose that [ b]≤[a] for any npositive numbers
x1,x2,…,xn.T a k i n g xi=x, for all i,w eo b t a i n
x/summationtextbi=[b]≤[a]=x/summationtextai.
T h i sc a no n l yb et r u ef o ra l l xif/summationtextbi=/summationtextai.
Next, take x1=x2=···=xν=x,xν+1=···=xn=1a n d xvery large. Since
(b)a n d( a) are in decreasing order, the index of the highest powers of xin [b]a n d
[a]a r e
b1+b2+···+bν,a1+a2+···+aν,
respectively. Thus, it is clear that the ﬁrst sum can not be greater than the second
and this proves (2) in Deﬁnition 1.8.1.
The proof in the other direction is more diﬃcult to establish, and we will
need a new deﬁnition and two more lemmas.
We deﬁne a special type of linear transformation Tof the a’s, as follows.
Suppose that ak>al, then let us write
ak=ρ+τ,al=ρ−τ(0<τ≤ρ).
If now 0 ≤σ<τ ≤ρ,t h e na T-transformation is deﬁned by
T(ak)=bk=ρ+σ=τ+σ
2τak+τ−σ
2τal,
T(al)=bl=ρ−σ=τ−σ
2τak+τ+σ
2τal,
T(aν)=aν(ν/negationslash=k,ν/negationslash=l).
If (b) arises from ( a)b ya T-transformation, we write b=Ta. The deﬁnition does
not necessarily imply that either the ( a)o rt h e( b) are in decreasing order.
The suﬃciency of our comparability condition will be established if we can prove
the following two lemmas.
Lemma 1.8.3. Ifb=Ta,t h e n [b]≤[a]with equality taking place only when all the
xi’s are equal.
Proof. We may rearrange ( a)a n d( b)s ot h a t k=1 ,l=2 .T h u s
[a]−[b]=[ρ+τ,ρ−τ,a 3,…]−[ρ+σ, ρ−σ, a 3,…]
=1
2n!/summationdisplay
!xa3
3···xan
n(xρ+τ
1xρ−τ
2+xρ−τ
1xρ+τ
2)
−1
2n!/summationdisplay
!xa3
3···xan
n(xρ+σ
1xρ−σ
2+xρ−σ
1xρ+σ
2)
=1
2n!/summationdisplay
!(x1x2)ρ−τxa3
3···xan
n(xτ+σ
1−xτ+σ
2)(xτ−σ
1−xτ−σ
2)≥0
with equality being the case only when all the xi’s are equal. /square

46 Numerical Inequalities
Lemma 1.8.4. If(b)≺(a),b u t(b)is not identical to (a),t h e n (b)can be derived
from(a)using the successive application of a ﬁnite number of T-transformations.
Proof. We call the number of diﬀerences aν−bνwhich are not zero, the discrepancy
between ( a)a n d( b). If the discrepancy is zero, the sets are identical. We will prove
the lemma by induction, assuming it to be true when the discrepancy is less than
rand proving that it is also true when the discrepancy is r.
Suppose then that ( b)≺(a) and that the discrepancy is r>0. Since/summationtextn
i=1ai=/summationtextni=1bi,a n d/summationtext(aν−bν) = 0, and not all of these diﬀerences are zero,
there must be positive and negative diﬀerences, and the ﬁrst which is not zero
must be positive because of the second condition of ( b)≺(a). We can therefore
ﬁndkandlsuch that
bk<ak,bk+1=ak+1,…,bl−1=al−1,bl>al; (1.12)
that is, al−blis the ﬁrst negative diﬀerence and ak−bkis the last positive
diﬀerence which precedes it.
We take ak=ρ+τ,al=ρ−τ, and deﬁne σby
σ=m a x ( |bk−ρ|,|bl−ρ|).
Then 0 <τ≤ρ,s i n c e ak>a l. Also, one (possible both) of bl−ρ=−σor
bk−ρ=σis true, since bk≥bl,a n d σ<τ ,s i n c e bk<akandbl>al. Hence
0≤σ<τ ≤ρ.
We now write a/prime
k=ρ+σ,a/primel=ρ−σ,a/prime
ν=aν(ν/negationslash=k,ν/negationslash=l). Ifbk−ρ=σ,
a/prime
k=bk,a n di f bl−ρ=−σ,t h e n a/primel=bl.S i n c et h ep a i r s ak,bkandal,bleach
contributes one unit to the discrepancy rbetween ( b)a n d( a), the discrepancy
between ( b)a n d( a/prime) is smaller, being equal to r−1o rr−2.
Next, comparing the deﬁnition of ( a/prime) with the deﬁnition of the T-transfor-
mation, and observing that 0 ≤σ<τ ≤ρ,w ec a ni n f e rt h a t( a/prime) arises from ( a)
by aT-transformation.
Finally, let us prove that ( b)≺(a/prime). In order to do that, we must verify that
the two conditions of ≺are satisﬁed and that the order of ( a/prime) is non-increasing.
For the ﬁrst one, we have
a/prime
k+a/primel=2ρ=ak+al,n/summationdisplay
i=1bi=n/summationdisplay
i=1ai=n/summationdisplay
i=1a/primei.
For the second one, we must prove that
b1+b2+···+bν≤α/prime1+α/prime2+···+α/primeν.
Now, this is true if ν<k orν≥l, as can be established by using the deﬁnition of
(a/prime) and also the second condition of ( b)≺(a). It is true for ν=k, because it is
true for ν=k−1a n d bk≤a/prime
k, and it is true for k<ν<l because it is valid for
ν=kand the intervening banda/primeare identical.

1.8 Muirhead’s theorem 47
Finally, we observe that
bk≤ρ+|bk−ρ|≤ρ+σ=a/prime
k,
bl≥ρ−|bl−ρ|≥ρ−σ=a/primel,
and then, using (1.12),
a/primek−1=ak−1≥ak=ρ+τ>ρ +σ=a/primek≥bk≥bk+1=ak+1=a/primek+1,
a/primel−1=al−1=bl−1≥bl≥a/primel=ρ−σ>ρ −τ=al≥al+1=a/primel+1.
The inequalities involving a/primeare as required.
We have thus proved that ( b)≺(a/prime), a set arising from ( a) using a transfor-
mation Tand having a discrepancy from ( b)o fl e s st h a n r. This proves the lemma
and completes the proof of Muirhead’s theorem. /square
The proof of Muirhead’s theorem demonstrates to us how the diﬀerence be-
tween two comparable means can be decomposed as a sum of obviously positive
terms by repeated application of the T-transformation. We can produce from this
result a new proof for the AM-GM inequality.
Example 1.8.5 (The AM-GM inequality). For real positive numbers y1,y2,…,
yn,
y1+y2+···+yn
n≥n√
y1y2···yn.
Note that the AM-GM inequality is equivalent to
1
nn/summationdisplay
i=1xn
i≥x1x2···xn,
where xi=n√
yi.
Now, we observe that
1
nn/summationdisplay
i=1xn
i=[n,0,0,…,0] and x1x2···xn=[ 1,1,…,1].
By Muirhead’s theorem we can show that
[n,0,0,…,0]≥[1,1,…,1].
Next, we provide another proof for the AM-GM inequality, something we
shall do by following the ideas inherent in the proof of Muirhead’s theorem in

48 Numerical Inequalities
order to illustrate how it works.
1
nn/summationdisplay
i=1xn
i−(x1x2···xn)=[n,0,0,…,0]−[1,1,…,1]
=( [n,0,0,…,0]−[n−1,1,0,…,0])
+( [n−1,1,0,…,0]−[n−2,1,1,0,…,0])
+( [n−2,1,1,0,…,0]−[n−3,1,1,1,0,…,0])
+···+( [ 2,1,1,…,1]−[1,1,…1])
=1
2n!/parenleftBig/summationdisplay
!(xn−1
1−xn−1
2)(x 1−x2)
+/summationdisplay
!(xn−2
1−xn−2
2)(x 1−x2)x3
+/summationdisplay
!(xn−3
1−xn−3
2)(x 1−x2)x3x4+···/parenrightBig
.
Since ( xν
r−xνs)(xr−xs)>0, unless xr=xs, the inequality follows.
Example 1.8.6. Ifa,bare positive real numbers, then
/radicalbigg
a2
b+/radicalbigg
b2
a≥√
a+√
b.
Setting x=√
a,y=√
band simplifying, we have to prove
x3+y3≥xy(x+y).
Using Muirhead’s theorem, we get
[3,0] =1
2(x3+y3)≥1
2xy(x+y)=[ 2 ,1],
and thus the result follows.
Example 1.8.7. Ifa,b,care non-negative real numbers, prove that
a3+b3+c3+abc≥1
7(a+b+c)3.
It is not diﬃcult to see that
(a+b+c)3=3 [ 3,0,0] + 18[2, 1,0] + 36[1, 1,1].
Then we need to prove that
3[3,0,0] + 6[1, 1,1]≥1
7(3[3,0,0] + 18[2, 1,0] + 36[1, 1,1]),
that is,
18
7[3,0,0] +/parenleftbigg
6−36
7/parenrightbigg
[1,1,1]≥18
7[2,1,0]

1.8 Muirhead’s theorem 49
or
18
7([3,0,0]−[2,1,0]) +/parenleftbigg
6−36
7/parenrightbigg
[1,1,1]≥0.
This follows using the inequalities [3 ,0,0]≥[2,1,0] and [1 ,1,1]≥0.
Example 1.8.8. Ifa,b,care non-negative real numbers, prove that
a+b+c≤a2+b2
2c+b2+c2
2a+c2+a2
2b≤a3
bc+b3
ca+c3
ab.
The inequalities are equivalent to the following:
2(a2bc+ab2c+abc2)≤ab(a2+b2)+bc(b2+c2)+ca(c2+a2)≤2(a4+b4+c4),
w h i c hi si nt u r ne q u i v a l e n tt o[ 2 ,1,1]≤[3,1,0]≤[4,0,0]. Using Muirhead’s theo-
rem we arrive at the result.
Exercise 1.115. Prove that any three positive real numbers a,bandcsatisfy
a5+b5+c5≥a3bc+b3ca+c3ab.
Exercise 1.116. (IMO, 1961) Let a,b,cbe the lengths of the sides of a triangle,
and let ( ABC) denote its area. Prove that
4√
3(ABC)≤a2+b2+c2.
Exercise 1.117. Leta,b,cbe positive real numbers. Prove that
a
(a+b)(a+c)+b
(b+c)(b+a)+c
(c+a)(c+b)≤9
4(a+b+c).
Exercise 1.118. (IMO, 1964) Let a,b,cbe positive real numbers. Prove that
a3+b3+c3+3abc≥ab(a+b)+bc(b+c)+ca(c+a).
Exercise 1.119. (Short list Iberoamerican, 2003) Let a,b,cbe positive real num-
bers. Prove that
a3
b2−bc+c2+b3
c2−ca+a2+c3
a2−ab+b2≥a+b+c.
Exercise 1.120. (Short list IMO, 1998) Let a,b,cbe positive real numbers such
thatabc=1 .P r o v et h a t
a3
(1 +b)(1 +c)+b3
(1 +c)(1 + a)+c3
(1 +a)(1 +b)≥3
4.

Chapter 2
Geometric Inequalities
2.1 Two basic inequalities
The two basic geometric inequalities we will be refering to in this section involve
triangles. One of them is the triangle inequality and we will refer to it as D1; the
second one is not really an inequality, but it represents an important observation
concerning the geometry of triangles which points out that if we know the greatest
angle of a triangle, then we know which is the longest side of the triangle; this
observation will be denoted as D2.
D1.IfA,BandCare points on the plane, then
AB+BC≥AC.
Moreover, the equality holds if and only if Blies on the line segment AC.
D2.In a triangle, the longest side is opposite to the greatest angle and vice versa.
Hence, if in the triangle ABC we have ∠A>∠B,t h e n BC > CA .
Exercise 2.1. (i) If a,b,care positive numbers with a<b +c,b<c +aand
c<a +b, then a triangle exists with side lengths a,bandc.
(ii) To be able to construct a triangle with side lengths a≤b≤c, it is suﬃcient
thatc<a +b.
(iii) It is possible to construct a triangle with sides of length a,bandcif and
only if there are positive numbers x,y,zsuch that a=x+y,b=y+zand
c=z+x.
Exercise 2.2. (i) If it is possible to construct a triangle with side-lengths a<b<
c, then it is possible to construct a triangle with side-lengths√
a<√
b<√
c.
(ii) The converse of (i) is false.

52 Geometric Inequalities
(iii) If it is possible to construct a triangle with side-lengths a<b<c ,t h e ni ti s
possible to construct a triangle with side-lengths1
a+b,1
b+cand1
c+a.
Exercise 2.3. Leta,b,c,dandebe the lengths of ﬁve segments such that it is
possible to construct a triangle using any three of them. Prove that there are three
of them that form an acute triangle.
Sometimes the key to solve a problem lies in the ability to identify certain
quantities that can be related to geometric measurements, as in the followingexample.
Example 2.1.1. Ifa,b,care positive numbers with a
2+b2−ab=c2, prove that
(a−b)(b−c)≤0.
Since c2=a2+b2−ab=a2+b2−2abcos60◦,w ec a nt h i n kt h a t a,b,care
the lengths of the sides of a triangle such that the measure of the angle opposed
to the side of length cis 60◦. The angles of the triangle ABC satisfy ∠A≤60◦
and∠B≥60◦,o r∠A≥60◦and∠B≤60◦; hence, using property D2 we can
deduce that a≤c≤bora≥c≥b. In any case it follows that (a −b)(b−c)≤0.
Observation 2.1.2. We can also solve the example above without the identiﬁcation
ofa,bandcwith the lengths of the sides of a triangle.
First suppose that a≤b, then the fact that a2+b2−ab=c2implies that
a(a−b)=c2−b2=(c−b)(c+b),h e n c e c−b≤0and therefore (a−b)(b−c)≤0.
Similarly, a≥bimplies c−b≥0, and hence
(a−b)(b−c)≤0.
Another situation where it is not obvious that we can identify the elements
with a geometric inequality, or that the use of geometry may be helpful, is shownin the following example.
Example 2.1.3. Ifa,b,care positive numbers, then
/radicalbig
a2+ac+c2≤/radicalbig
a2−ab+b2+/radicalbig
b2−bc+c2.
The radicals suggest using the cosine law with angles of 120◦and of 60◦as
follows: a2+ac+c2=a2+c2−2accos120◦,a2−ab+b2=a2+b2−2abcos60◦
andb2−bc+c2=b2+c2−2bccos60◦.
A
B CD
ba
c60◦60◦

2.1 Two basic inequalities 53
Then, if we consider a quadrilateral ABCD ,w i t h ∠ADB =∠BDC =6 0◦and
∠ADC = 120◦, such that AD=a,BD=bandCD=c, we can deduce that AB=√
a2−ab+b2,BC=√
b2−bc+c2andCA=√
a2+ac+c2. The inequality we
have to prove becomes the triangle inequality for the triangle ABC.
Exercise 2.4. LetABC be a triangle with ∠A>∠B,p r o v et h a t BC >1
2AB.
Exercise 2.5. LetABCD be a convex quadrilateral, prove that
(i) ifAB+BD < AC +CD,t h e n AB < AC ,
(ii) if∠A>∠Cand∠D>∠B,t h e n BC >1
2AD.
Exercise 2.6. Ifa1,a2,a3,a4anda5are the lengths of the sides of a convex
pentagon and if d1,d2,d3,d4andd5are the lengths of its diagonals, prove that
1
2<a1+a2+a3+a4+a5
d1+d2+d3+d4+d5<1.
Exercise 2.7. The length maof the median AA/primeof a triangle ABC satisﬁes ma>
b+c−a
2.
Exercise 2.8. If the length maof the median AA/primeof a triangle ABC satisﬁes
ma>1
2a,p r o v et h a t ∠BAC < 90◦.
Exercise 2.9. IfAA/primeis the median of the triangle ABC and if AB < AC ,t h e n
∠BAA/prime>∠A/primeAC.
Exercise 2.10. Ifma,mbandmcare the lengths of the medians of a triangle with
side-lengths a,bandc, respectively, prove that it is possible to construct a triangle
with side-lengths ma,mbandmc,a n dt h a t
3
4(a+b+c)<m a+mb+mc<a+b+c.
Exercise 2.11. (Ptolemy’s inequality) IfABCD is a convex quadrilateral, then
AC·BD≤AB·CD+BC·DA. The equality holds if and only if ABCD is a
cyclic quadrilateral.
Exercise 2.12. LetABCD be a cyclic quadrilateral. Prove that AC > BD if and
only if ( AD−BC)(AB−DC)>0.
Exercise 2.13. (Pompeiu’s problem) LetABC be an equilateral triangle and let
Pbe a point that does not belong to the circumcircle of ABC.P r o v et h a t PA,
PBandPCare the lengths of the sides of a triangle.
Exercise 2.14. IfABCD is a paralelogram, prove that
|AB2−BC2|<A C ·BD.

54 Geometric Inequalities
Exercise 2.15. Ifa,bandcare the lengths of the sides of a triangle, ma,mband
mcrepresent the lengths of the medians and Ris the circumradius, prove that
(i)a2+b2
mc+b2+c2
ma+c2+a2
mb≤12R,
(ii)ma(bc−a2)+mb(ca−b2)+mc(ab−c2)≥0.
Exercise 2.16. LetABC be a triangle whose sides have lengths a,bandc. Suppose
thatc>b,p r o v et h a t
1
2(c−b)<m b−mc<3
2(c−b),
where mbandmcare the lengths of the medians.
Exercise 2.17. (Iran, 2005) Let ABC be a triangle with ∠A=9 0◦.L e tDbe the
intersection of the internal angle bisector of ∠Awith the side BCand let Iabe
the center of the excircle of the triangle ABC opposite to the vertex A.P r o v et h a t
AD
DIa≤√
2−1.
2.2 Inequalities between the sides of a triangle
Inequalities involving the lengths of the s ides of a triangle appear frequently in
mathematical competitions. One sort of problems consists of those where you are
asked to prove some inequality that is satisﬁed by the lengths of the sides of a
triangle without any other geometric elements being involved, as in the followingexample.
Example 2.2.1. The lengths a,bandcof the sides of a triangle satisfy
a(b+c−a)<2bc.
Since the inequality is symmetric in bandc, we can assume, without loss of
generality, that c≤b. We will prove the inequality in the following cases.
Case 1. a≤b.
Since they are the lengths of the sides of a triangle, we have that b<a +c;t h e n
b+c−a=b−a+c<c+c=2c≤2bc
a.
Case 2. a≥b.
In this case b−a≤0, and since a<b +c≤2b, we can deduce that
b+c−a=c+b−a≤c<2bc
a.
Another type of problem involving the lengths of the sides of a triangle is
when we are asked to prove that a certain relationship between the numbers a,b
andcis suﬃcient to construct a triangle with sides of the same length.

2.2 Inequalities between the sides of a triangle 55
Example 2.2.2. (i)Ifa,b,care positive numbers and satisfy,/parenleftbig
a2+b2+c2/parenrightbig2>
2/parenleftbig
a4+b4+c4/parenrightbig
,t h e n a,bandcare the lengths of the sides of a triangle.
(ii)Ifa,b,c,dare positive numbers and satisfy
/parenleftbig
a2+b2+c2+d2/parenrightbig2>3/parenleftbig
a4+b4+c4+d4/parenrightbig
,
then,using any three of them we can construct a triangle.
F o rp a r t( i ) ,i ti ss u ﬃ c i e n tt oo b s e r v et h a t
/parenleftbig
a2+b2+c2/parenrightbig2−2/parenleftbig
a4+b4+c4/parenrightbig
=(a+b+c)(a+b−c)(a−b+c)(−a+b+c)>0,
and then note that none of these factors is negative. Compare this with Example
1.2.5.
For part (ii), we can deduce that
3/parenleftbig
a4+b4+c4+d4/parenrightbig
</parenleftbig
a2+b2+c2+d2/parenrightbig2
=/parenleftbigga2+b2+c2
2+a2+b2+c2
2+d2/parenrightbigg2
≤/braceleftBigg/parenleftbigga2+b2+c2
2/parenrightbigg2
+/parenleftbigga2+b2+c2
2/parenrightbigg2
+d4/bracerightBigg/parenleftBig√
3/parenrightBig2
.
The second inequality follows from the Cauchy-Schwarz inequality; hence, a4+
b4+c4<2(a2+b2+c2)
42
. Using the ﬁrst part we can deduce that a,bandccan be
used to construct a triangle. Since the argument we used is symmetric in a,b,c
andd, we obtain the result.
There is a technique that helps to transform one inequality between the
lengths of the sides of a triangle into an inequality between positive numbers (ofcourse related to the sides).This is called the Ravi transformation .
If the incircle ( I,r) of the triangle ABC is tangent to the sides BC,CA
andABat the points X,YandZ, respectively, we have that x=AZ=YA,
y=ZB=BXandz=XC=CY.
/Bullet
/Bullet/Bullet
/Bullet
BCA
I
XY
Zxx
y
y zz
It is easily seen that a=y+z,b=z+x,c=x+y,x=s−a,y=s−band
z=s−c,w h e r e s=a+b+c
2.
Let us now see how to use the Ravi transformation in the following example.

56 Geometric Inequalities
Example 2.2.3. The lengths of the sides a,bandcof a triangle satisfy
(b+c−a)(c+a−b)(a+b−c)≤abc.
F i r s t ,w eh a v et h a t
(b+c−a)(c+a−b)(a+b−c)=8 ( s−a)(s−b)(s−c)=8xyz,
on the other hand
abc=(x+y)(y+z)(z+x).
Thus, the inequality is equivalent to
8xyz≤(x+y)(y+z)(z+x). (2.1)
The last inequality follows from Exercise 1.26.
Example 2.2.4. (APMO, 1996) Leta,b,cbe the lengths of the sides of a triangle,
prove that√
a+b−c+√
b+c−a+√
c+a−b≤√
a+√
b+√
c.
If we set a=y+z,b=z+x,c=x+y, we can deduce that a+b−c=2z,
b+c−a=2x,c+a−b=2y. Hence, the inequality is equivalent to
√
2x+/radicalbig
2y+√
2z≤√
x+y+√
y+z+√
z+x.
Now applying the inequality between the arithmetic mean and the quadratic mean
(see Exercise 1.68), we get
√
2x+/radicalbig
2y+√
2z=√
2x+√
2y
2+√
2y+√
2z
2+√
2z+√
2x
2
≤/radicalbigg
2x+2y
2+/radicalbigg
2y+2z
2+/radicalbigg
2z+2x
2
=√
x+y+√
y+z+√
z+x.
Moreover, the equality holds if and only if x=y=z, that is, if and only if
a=b=c.
Also, it is possible to express the area of a triangle ABC, its inradius, its
circumradius and its semiperimeter in terms of x,y,z.S i n c e a=x+y,b=y+z
andc=z+x, we ﬁrst obtain that s=a+b+c
2=x+y+z. Using Heron’s formula
for the area of a triangle, we get
(ABC)=/radicalbig
s(s−a)(s−b)(s−c)=/radicalbig
(x+y+z)xyz. (2.2)
The formula ( ABC)=srleads us to
r=(ABC)
s=/radicalbig
(x+y+z)xyz
x+y+z=/radicalbigg
xyz
x+y+z.

2.2 Inequalities between the sides of a triangle 57
Finally, from ( ABC)=abc
4Rwe get
R=(x+y)(y+z)(z+x)
4/radicalbig
(x+y+z)xyz.
Example 2.2.5. (India, 2003) Leta,b,cbe the side lengths of a triangle ABC.I f
we construct a triangle A/primeB/primeC/primewith side lengths a+b
2,b+c
2,c+a
2, prove that
(A/primeB/primeC/prime)≥9
4(ABC).
Since a=y+z,b=z+xandc=x+y, the side lengths of the triangle
A/primeB/primeC/primearea/prime=x+2y+3z
2,b/prime=3x+y+2z
2,c/prime=2x+3y+z
2. Using Heron’s formula for
the area of a triangle, we get
(A/primeB/primeC/prime)=/radicalbigg
3(x+y+z)(2x+y)(2y+z)(2z+x)
16.
Applying the AM-GMinequality to show that 2 x+y≥33/radicalbig
x2y,2y+z≥33/radicalbig
y2z,
2z+x≥33√
z2x, will help to reach the inequality
(A/primeB/primeC/prime)≥/radicalbigg
3(x+y+z)27(xyz)
16=9
4(ABC).
Equation (2.2) establishes the last equality.
Exercise 2.18. Leta,bandcbe the lengths of the sides of a triangle, prove that
3(ab+bc+ca)≤(a+b+c)2≤4(ab+bc+ca).
Exercise 2.19. Leta,bandcbe the lengths of the sides of a triangle, prove that
ab+bc+ca≤a2+b2+c2≤2(ab+bc+ca).
Exercise 2.20. Leta,bandcbe the lengths of the sides of a triangle, prove that
2/parenleftbig
a2+b2+c2/parenrightbig
≤(a+b+c)2.
Exercise 2.21. Leta,bandcbe the lengths of the sides of a triangle, prove that
3
2≤a
b+c+b
c+a+c
a+b<2.
Exercise 2.22. (IMO, 1964) Let a,bandcbe the lengths of the sides of a triangle,
prove that
a2(b+c−a)+b2(c+a−b)+c2(a+b−c)≤3abc.

58 Geometric Inequalities
Exercise 2.23. Leta,bandcbe the lengths of the sides of a triangle, prove that
a/parenleftbig
b2+c2−a2/parenrightbig
+b/parenleftbig
c2+a2−b2/parenrightbig
+c/parenleftbig
a2+b2−c2/parenrightbig
≤3abc.
Exercise 2.24. (IMO, 1983) Let a,bandcbe the lengths of the sides of a triangle,
prove that
a2b(a−b)+b2c(b−c)+c2a(c−a)≥0.
Exercise 2.25. Leta,bandcbe the lengths of the sides of a triangle, prove that
/vextendsingle/vextendsingle/vextendsingle/vextendsinglea−b
a+b+b−c
b+c+c−a
c+a/vextendsingle/vextendsingle/vextendsingle/vextendsingle<1
8.
Exercise 2.26. The lengths a,bandcof the sides of a triangle satisfy ab+bc+ca=3 .
Prove that
3≤a+b+c≤2√
3.
Exercise 2.27. Leta,b,cbe the lengths of the sides of a triangle, and let rbe the
inradius of the triangle. Prove that
1
a+1
b+1
c≤√
3
2r.
Exercise 2.28. Leta,b,cbe the lengths of the sides of a triangle, and let sbe the
semiperimeter of the triangle. Prove that
(i) (s−a)(s−b)<a b,
(ii) (s−a)(s−b)+(s−b)(s−c)+(s−c)(s−a)≤ab+bc+ca
4.
Exercise 2.29. Ifa,b,care the lengths of the sides of an acute triangle, prove that
/summationdisplay
cyclic/radicalbig
a2+b2−c2/radicalbig
a2−b2+c2≤a2+b2+c2,
where/summationtext
cyclicstands for the sum over all cyclic permutations of ( a,b,c).
Exercise 2.30. Ifa,b,care the lengths of the sides of an acute triangle, prove that
/summationdisplay
cyclic/radicalbig
a2+b2−c2/radicalbig
a2−b2+c2≤ab+bc+ca,
where/summationtext
cyclicrepresents the sum over all cyclic permutations of ( a,b,c).

2.3 The use of inequalities in the geometry of the triangle 59
2.3 The use of inequalities in the geometry of the
triangle
A problem which shows the use of inequalities in the geometry of the triangle was
introduced in the International Mathematical Olympiad in 1961; for this problem
there are several proofs and its applications are very broad, as will be seen later
on. Meanwhile, we present it here as an example.
Example 2.3.1. Ifa,bandcare the lengths of the sides of a triangle with area
(ABC),t h e n 4√
3(ABC)≤a2+b2+c2.
Since an equilateral triangle of side-length ahas area equal to√
3
4a2,t h e
equality in the example holds for this case; hence we will try to compare what
happens in any triangle with what happens in an equilateral triangle of side lengtha.
BCA
Dd eb c
h
LetBC=a.I fADis the altitude of the triangle at A, its length hcan be expressed
ash=√
3
2a+y,w h e r e ymeasures its diﬀerence in comparison with the length
of the altitude of the equilateral triangle. We also set d=a
2−xande=a
2+x,
where xcan be interpreted as the diﬀerence that the projection of AonBChas
with respect to the projection of AonBCin an equilateral triangle, which in this
case is the midpoint of BC.W eo b t a i n
a2+b2+c2−4√
3(ABC)=a2+h2+/parenleftBiga
2+x/parenrightBig2
+h2+/parenleftBiga
2−x/parenrightBig2
−4√
3ah
2
=3
2a2+2h2+2×2−2√
3a/parenleftBigg√
3
2a+y/parenrightBigg
=3
2a2+2/parenleftBigg√
3
2a+y/parenrightBigg2
+2×2−3a2−2√
3ay
=3
2a2+3
2a2+2√
3ay+2y2+2×2−3a2−2√
3ay
=2 (x2+y2)≥0.
Moreover, the equality holds if and only if x=y= 0, that is, when the triangle is
equilateral.

60 Geometric Inequalities
Let us give another proof for the previous example. Let ABC be a triangle,
with side-lengths a≥b≥c,a n dl e t A/primebe a point such that A/primeBCis an equilateral
triangle with side-length a.I fw et a k e d=AA/prime,t h e n dmeasures, in a manner,
how far is ABC from being an equilateral triangle.
AA/prime
BCabcad
Using the cosine law we can deduce that
d2=a2+c2−2accos(B−60◦)
=a2+c2−2ac(cosBcos60◦+s i nBsin60◦)
=a2+c2−accosB−2√
3acsinB
2
=a2+c2−ac/parenleftbigga2+c2−b2
2ac/parenrightbigg
−2√
3(ABC)
=a2+b2+c2
2−2√
3(ABC).
Butd2≥0, hence we can deduce that 4√
3(ABC)≤a2+b2+c2,w h i c hi sw h a t
we wanted to prove. Moreover, the equality holds if d=0 ,t h a ti s ,i f A/prime=Aor,
equivalently, if ABC is equilateral.
It is quite common to ﬁnd inequalities that involve elements of the triangle
among mathematical olympiad problems. Some of them are based on the following
inequality, which is valid for positive numbers a,b,c(see Exercise 1.36 of Section
1.3):
(a+b+c)/parenleftbigg1
a+1
b+1
c/parenrightbigg
≥9. (2.3)
Moreover, we recall that the equality holds if and only if a=b=c.
Another inequality, which has been very helpful to solve geometric-related
problems, is Nesbitt’s inequality (see Example 1.4.8 of Section 1.4). It states that
fora,b,cpositive numbers, we always have
a
b+c+b
c+a+c
a+b≥3
2. (2.4)

2.3 The use of inequalities in the geometry of the triangle 61
The previous inequality can be proved using inequality (2.3) as follows:
a
b+c+b
c+a+c
a+b=a+b+c
b+c+a+b+c
c+a+a+b+c
a+b−3
=(a+b+c)/parenleftbigg1
b+c+1
c+a+1
a+b/parenrightbigg
−3
=1
2[(a+b)+(b+c)+(c+a)]·/parenleftbigg1
b+c+1
c+a+1
a+b/parenrightbigg
−3
≥9
2−3=3
2.
The equality holds if and only if a+b=b+c=c+a, or equivalently, if a=b=c.
Let us now observe some examples of geometric inequalities where such re-
lationships are employed.
Example 2.3.2. LetABC be an equilateral triangle of side length a,l e tMbe a
point inside ABC and let D,E,Fbe the projections of Mon the sides BC,CA
andAB, respectively. Prove that
(i)1
MD+1
ME+1
MF≥6√
3
a,
(ii)1
MD+ME+1
ME+MF+1
MF+MD≥3√
3
a.
/BulletA
BCE F
DM
Letx=MD,y=MEandz=MF. Remember that we denote the area of
the triangle ABC as (ABC), then ( ABC)=(BCM )+(CAM )+(ABM ), hence
ah=ax+ay+az,w h e r e h=√
3
2arepresents the length of the altitude of ABC.
Therefore, h=x+y+z. (This result is known as Viviani’s lemma; see Section
2.8). Using inequality (2.3) we can deduce that
h/parenleftbigg1
x+1
y+1
z/parenrightbigg
≥9 and, after solving, that1
x+1
y+1
z≥9
h=6√
3
a.

62 Geometric Inequalities
To prove the second part, using inequality (2.3), we can establish that
(x+y+y+z+z+x)/parenleftbigg1
x+y+1
y+z+1
z+x/parenrightbigg
≥9.
Therefore,1
x+y+1
y+z+1
z+x≥9
2h=3√
3
a.
Example 2.3.3. Ifha,hbandhcare the lengths of the altitudes of the triangle
ABC, whose incircle has center Iand radius r, we have
(i)r
ha+r
hb+r
hc=1,
(ii)ha+hb+hc≥9r.
In order to prove the ﬁrst equation, observe thatr
ha=r·a
ha·a=(IBC )
(ABC ). Simi-
larly,r
hb=(ICA )
(ABC ),r
hc=(IAB )
(ABC ). Adding the three equations, we have that
r
ha+r
hb+r
hc=(IBC)
(ABC)+(ICA)
(ABC)+(IAB)
(ABC)
=(IBC)+(ICA)+(IAB)
(ABC)=1.
A
BCI
rha
The desired inequality is a straightforward consequence of inequality (2.3), since
(ha+hb+hc)/parenleftBig
1
ha+1
hb+1
hc/parenrightBig
·r≥9r.
Example 2.3.4. LetABC b eat r i a n g l ew i t ha l t i t u d e s AD,BE,CFand let Hbe
its orthocenter. Prove that
(i)AD
HD+BE
HE+CF
HF≥9,
(ii)HD
HA+HE
HB+HF
HC≥3
2.

2.3 The use of inequalities in the geometry of the triangle 63
A
BCHEF
D
To prove part (i), consider S=(ABC),S1=(HBC ),S2=(HCA),S3=
(HAB). Since triangles ABC andHBC share the same base, their area ratio is
equal to their altitude ratio, that is,S1
S=HD
AD. Similarly,S2
S=HE
BEandS3
S=HF
CF.
Then,HD
AD+HE
BE+HF
CF=1.
Using inequality (2.3) we can state that
/parenleftbiggAD
HD+BE
HE+CF
HF/parenrightbigg/parenleftbiggHD
AD+HE
BE+HF
CF/parenrightbigg
≥9.
If we substitute the equality previously calculated, we get (i).
Moreover, the equality holds if and only ifHD
AD=HE
BE=HF
CF=1
3,t h a ti s ,
ifS1=S2=S3=1
3S. To prove the second part observe thatHD
HA=HD
AD−HD=
S1
S−S1=S1
S2+S3, and similarly,HE
HB=S2
S3+S1andHF
HC=S3
S1+S2, then using Nesbitt’s
inequality leads toHD
HA+HE
HB+HF
HC≥3
2.
Example 2.3.5. (Korea, 1995) LetABC be a triangle and let L,M,Nbe points
onBC,CAandAB, respectively. Let P,QandRbe the intersection points of
the lines AL,BMandCNwith the circumcircle of ABC, respectively. Prove that
AL
LP+BM
MQ+CN
NR≥9.
LetA/primebe the midpoint of BC,l e tP/primebe the midpoint of the arc BC,l e tD
andD/primebe the projections of AandPonBC, respectively.
It is clear thatAL
LP=AD
PD/prime≥AD
P/primeA/prime. Thus, the minimum value ofAL
LP+BM
MQ+CN
NR
is attained when P,QandRare the midpoints of the arcs BC,CAandAB.T h i s
happens when AL,BMandCNare the internal angle bisectors of the triangle
ABC. Hence, without loss of generality, we will assume that AL,BMandCN
are the internal angle bisectors of ABC. Since ALis an internal angle bisector, we
have6
BL=ca
b+c,LC=ba
b+cand AL2=bc/parenleftBigg
1−/parenleftbigga
b+c/parenrightbigg2/parenrightBigg
.
6See [6, pages 74 and 105] or [9, pages 10,11].

64 Geometric Inequalities
A
BCA/prime
L
PD
P/primeD/prime
Moreover,
AL
LP=AL2
AL·LP=AL2
BL·LC=(bc)/parenleftbigg
1−/parenleftBig
a
b+c/parenrightBig2/parenrightbigg
a2bc
(b+c)2=(b+c)2−a2
a2.
Similarly, for the internal angle bisectors BMandCN,w eh a v e
BM
MQ=(c+a)2−b2
b2andCN
NR=(a+b)2−c2
c2.
Therefore,
AL
LP+BM
MQ+CN
NR=/parenleftbiggb+c
a/parenrightbigg2
+/parenleftbiggc+a
b/parenrightbigg2
+/parenleftbigga+b
c/parenrightbigg2
−3
≥1
3/parenleftbiggb+c
a+c+a
b+a+b
c/parenrightbigg2
−3
≥1
3(6)2−3=9.
The ﬁrst inequality follows from the convexity of the function f(x)=x2and the
second inequality from relations in the forma
b+b
a≥2. Observe that equality
holds if and only if a=b=c.
Another way to ﬁnish the problem is the following:
/parenleftbiggb+c
a/parenrightbigg2
+/parenleftbiggc+a
b/parenrightbigg2
+/parenleftbigga+b
c/parenrightbigg2
−3
=/parenleftbigga2
b2+b2
a2/parenrightbigg
+/parenleftbiggb2
c2+c2
b2/parenrightbigg
+/parenleftbiggc2
a2+a2
c2/parenrightbigg
+2/parenleftbiggab
c2+bc
a2+ca
b2/parenrightbigg
−3
≥2·3+2· 3−3=9.

2.3 The use of inequalities in the geometry of the triangle 65
Here we made use of the fact thata2
b2+b2
a2≥2a n dt h a tab
c2+bc
a2+ca
b2≥
33/radicalBig
(ab)(bc)(ca)
a2b2c2=3.
Example 2.3.6. (Shortlist IMO, 1997) The lengths of the sides of the hexagon
ABCDEF satisfy AB=BC,CD=DEandEF=FA. Prove that
BC
BE+DE
DA+FA
FC≥3
2.
AB
C
F
EDa
bc
Seta=AC,b=CEandc=EA. Ptolemy’s inequality (see Exercise 2.11), applied
to the quadrilateral ACEF , guarantees that AE·FC≤FA·CE+AC·EF.S i n c e
EF=FA,w eh a v et h a t c·FC≤FA·b+FA·a.Therefore,
FA
FC≥c
a+b.
Similarly, we can deduce the inequalities
BC
BE≥a
b+candDE
DA≥b
c+a.
Hence,BC
BE+DE
DA+FA
FC≥a
b+c+b
c+a+c
a+b≥3
2; the last inequality is Nesbitt’s
inequality.
Exercise 2.31. Leta,b,cbe the lengths of the sides of a triangle, prove that:
(i)a
b+c−a+b
c+a−b+c
a+b−c≥3,
(ii)b+c−a
a+c+a−b
b+a+b−c
c≥3.
Exercise 2.32. LetAD,BE,CFbe the altitudes of the triangle ABC and let PQ,
PR,PSbe the distances from a point Pto the sides BC,CA,AB, respectively.
Prove thatAD
PQ+BE
PR+CF
PS≥9.

66 Geometric Inequalities
Exercise 2.33. Through a point Oinside a triangle of area Sthree lines are drawn
in such a way that every side of the triangle intersects two of them. These lines
divide the triangle into three triangles with common vertex Oand areas S1,S2
andS3, and three quadrilaterals. Prove that
(i)1
S1+1
S2+1
S3≥9
S,
(ii)1
S1+1
S2+1
S3≥18
S.
Exercise 2.34. The cevians AL,BMandCNof the triangle ABC concur in P.
Prove thatAP
PL+BP
PM+CP
PN=6i fa n do n l yi f Pis the centroid of the triangle.
Exercise 2.35. The altitudes AD,BE,CFintersect the circumcircle of the triangle
ABC inD/prime,E/primeandF/prime, respectively. Prove that
(i)AD
DD/prime+BE
EE/prime+CF
FF/prime≥9,
(ii)AD
AD/prime+BE
BE/prime+CE
CF/prime≥9
4.
Exercise 2.36. In the triangle ABC,l e tla,lb,lcbe the lengths of the internal
bisectors of the angles of the triangle, and let sandrbe the semiperimeter and
the inradius of ABC.P r o v et h a t
(i)lalblc≤rs2,
(ii)lalb+lblc+lcla≤s2,
(iii)l2
a+l2
b+l2
c≤s2.
Exercise 2.37. LetABC be a triangle and let M,N,Pbe arbitrary points on the
line segments BC,CA,AB, respectively. Denote the lengths of the sides of the
triangle by a,b,cand the circumradius by R.P r o v et h a t
bc
AM+ca
BN+ab
CP≤6R.
Exercise 2.38. LetABC be a triangle with side-lengths a,b,c.L e tma,mband
mcbe the lengths of the medians from A,BandC, respectively. Prove that
max{ama,bm b,cm c}≤sR,
where Ris the radius of the circumcircle and sis the semiperimeter.
2.4 Euler’s inequality and some applications
Theorem 2.4.1 (Euler’s theorem). Given the triangle ABC,w h e r e Ois the cir-
cumcenter, Ithe incenter, Rthe circumradius and rthe inradius, then
OI2=R2−2Rr.

2.4 Euler’s inequality and some applications 67
Proof. Let us give a proof7that depends only on Pythagoras theorem and the fact
that the circumcircle of the triangle BCIhas center D, the midpoint of the arc8
BC. For the proof we will use directed segments.
/Bullet
/Bullet/BulletA
BC
DMOQI
LetMbe the midpoint of BCand let Qbe the orthogonal projection of Ion the
radius OD.T h e n
OB2−OI2=OB2−DB2+DI2−OI2
=OM2−MD2+DQ2−QO2
=(MO+DM)(MO−DM)+(DQ+QO)(DQ−QO)
=DO(MO+MD+DQ+OQ)
=R(2MQ)=2Rr.
Therefore OI2=R2−2Rr. /square
As a consequence of the last theorem w e obtain the following inequality.
Theorem 2.4.2 (Euler’s inequality). R≥2r. Moreover, R=2rif and only if the
triangle is equilateral.9
7Another proof can be found in [6, page 122] or [9, page 29].
8The proof can be found in [6, observation 3.2.7, page 123] or [1, page 76].
9There are direct proofs for the inequality (that is, without having to use Euler’s formula).
One of them is the following: the nine-point circle of a triangle is the circumcircle of the medial
triangle A/primeB/primeC/prime. Because this triangle is similar to ABC with ratio 2:1, we can deduce that the
radius of the nine-point circle isR
2.Clearly, a circle that intersects the three sides of a triangle
must have a greater radius than the radius of the incircle, thereforeR
2≥r.

68 Geometric Inequalities
Theorem 2.4.3. In a triangle ABC, with circumradius R, inradius rand semipe-
rimeter s, it happens that
r≤s
3√
3≤R
2.
Proof. We will use the fact that10(ABC)=abc
4R=sr.U s i n gt h e AM-GM
inequality, we can deduce that 2 s=a+b+c≥33√
abc=33√
4Rrs.T h u s ,
8s3≥27(4Rrs )≥27(8r2s), since R≥2r.Therefore, s≥3√
3r.
The second inequality,s
3√
3≤R
2,i se q u i v a l e n tt o a+b+c≤3√
3R. But using
the sine law, this is equivalent to sin A+s i nB+s i nC≤3√
3
2.Observe that the
last inequality holds because the function f(x)=s i n xis concave on [0 ,π], thus
sinA+sinB+sinC
3≤sin/parenleftbigA+B+C
3/parenrightbig
=s i n6 0◦=√
3
2. /square
Exercise 2.39. Leta,bandcbe the lengths of the sides of a triangle, prove that
(a+b−c)(b+c−a)(c+a−b)≤abc.
Exercise 2.40. Leta,bandcbe the lengths of the sides of a triangle, prove that
1
ab+1
bc+1
ca≥1
R2,
where Rdenotes the circumradius.
Exercise 2.41. LetA,BandCbe the measurements of the angles in each of the
vertices of the triangle ABC,p r o v et h a t
1
sinAsinB+1
sinBsinC+1
sinCsinA≥4.
Exercise 2.42. LetA,BandCbe the measurements of the angles in each of the
vertices of the triangle ABC,p r o v et h a t
/parenleftbigg
sinA
2/parenrightbigg/parenleftbigg
sinB
2/parenrightbigg/parenleftbigg
sinC
2/parenrightbigg
≤1
8.
Exercise 2.43. LetABC be a triangle. Call A,BandCthe angles in the vertices
A,BandC, respectively. Let a,bandcbe the lengths of the sides of the triangle
and let Rbe the radius of the circumcircle. Prove that
/parenleftbigg2A
π/parenrightbigg1
a/parenleftbigg2B
π/parenrightbigg1
b/parenleftbigg2C
π/parenrightbigg1
c
≤/parenleftbigg2
3/parenrightbigg√
3
R
.
Theorem 2.4.4 (Leibniz’s theorem). In a triangle ABC with sides of length a,b
andc, and with circumcenter O,c e n t r o i d Gand circumradius R, the following
holds:
OG2=R2−1
9/parenleftbig
a2+b2+c2/parenrightbig
.
10See [6, page 97] or [9, page 13].

2.4 Euler’s inequality and some applications 69
Proof. Let us use Stewart’s theorem which states11that if Lis a point on the side
BCof a triangle ABC and if AL=l,BL=m,LC=n,t h e n a/parenleftbig
l2+mn/parenrightbig
=
b2m+c2n.
/Bullet/BulletA
BC
A/primeGO
R
Applying Stewart’s theorem to the triangle OAA/primeto ﬁnd the length of OG,w h e r e
A/primeis the midpoint of BC,w eg e t
AA/prime/parenleftbig
OG2+AG·GA/prime/parenrightbig
=A/primeO2·AG+AO2·GA/prime.
Since
AO=R, AG =2
3AA/primeand GA/prime=1
3AA/prime,
substituting we get
OG2+2
9(A/primeA)2=A/primeO2·2
3+R2·1
3.
On the other hand12,s i n c e( A/primeA)2=2(b2+c2)−a2
4andA/primeO2=R2−a2
4,w ec a n
deduce that
OG2=/parenleftbigg
R2−a2
4/parenrightbigg2
3+1
3R2−2
9/parenleftBigg
2/parenleftbig
b2+c2/parenrightbig
−a2
4/parenrightBigg
=R2−a2
6−2/parenleftbig
b2+c2/parenrightbig
−a2
18
=R2−a2+b2+c2
9.
/square
One consequence of the last theorem is the following inequality.
Theorem 2.4.5 (Leibniz’s nequality). In a triangle ABCwith side-lengths a,band
c, with circumradius R, the following holds:
9R2≥a2+b2+c2.
11For a proof see [6, page 96] or [9, page 6].
12See [6, page 83] or [9, page 10].

70 Geometric Inequalities
Moreover, equality holds if and only if O=G, that is, when the triangle is equi-
lateral.
Example 2.4.6. In a triangle ABC with sides of length a,bandc, it follows that
4√
3(ABC)≤9abc
a+b+c.
Using that 4 R(ABC)=abc, we have the following equivalences:
9R2≥a2+b2+c2⇔a2b2c2
16(ABC)2≥a2+b2+c2
9⇔4(ABC)≤3abc
√
a2+b2+c2.
Cauchy-Schwarz inequality says that a+b+c≤√
3√
a2+b2+c2, hence
4√
3(ABC)≤9abc
a+b+c.
Exercise 2.44. LetA,BandCbe the measurements of the angles in each of the
vertices of the triangle ABC,p r o v et h a t
sin2A+s i n2B+s i n2C≤9
4.
Exercise 2.45. Leta,bandcbe the lengths of the sides of a triangle, prove that
4√
3(ABC)≤33√
a2b2c2.
Exercise 2.46. Suppose that the incircle of ABC is tangent to the sides BC,CA,
AB,a tD,E,F, respectively. Prove that
EF2+FD2+DE2≤s2
3,
where sis the semiperimeter of ABC.
Exercise 2.47. Leta,b,cbe the lenghts of the sides of a triangle ABC and let ha,
hb,hcbe the lenghts of the altitudes over BC,CA,AB, respectively. Prove that
a2
hbhc+b2
hcha+c2
hahb≥4.
2.5 Symmetric functions of a,bandc
The lengths of the sides a,bandcof a triangle have a very close relationship with
s,randR, the semiperimeter, the inradius and the circumradius of the triangle,
respectively. The relationships that are most commonly used are
a+b+c=2s, (2.5)
ab+bc+ca=s2+r2+4rR, (2.6)
abc=4Rrs. (2.7)

2.5 Symmetric functions of a,bandc 71
The ﬁrst is the deﬁnition of sand the third follows from the fact that the area of
the triangle isabc
4R=rs.Using Heron’s formula for the area of a triangle, we have
the relationship s(s−a)(s−b)(s−c)=r2s2, hence
s3−(a+b+c)s2+(ab+bc+ca)s−abc=r2s.
If we substitute (2.5) and (2.7) in this equality, after simplifying we get that
ab+bc+ca=s2+r2+4Rr.
Now, since any symmetric polynomial in a,bandccan be expressed as a polyno-
mial in terms of ( a+b+c), (ab+bc+ca)a n d( abc), it can also be expressed as a
polynomial in s,randR. For instance,
a2+b2+c2=(a+b+c)2−2(ab+bc+ca)=2/parenleftbig
s2−r2−4Rr/parenrightbig
,
a3+b3+c3=(a+b+c)3−3(a+b+c)(ab+bc+ca)+3 abc
=2/parenleftbig
s3−3r2s−6Rrs/parenrightbig
.
These transformations help to solve diﬀerent problems, as will be seen later
on.
Lemma 2.5.1. IfA,BandCare the measurements of the angles within each of
the vertices of the triangle ABC, we have that cosA+c o s B+c o s C=r
R+1.
Proof.
cosA+c o s B+c o s C=b2+c2−a2
2bc+c2+a2−b2
2ca+a2+b2−c2
2ab
=a/parenleftbig
b2+c2/parenrightbig
+b/parenleftbig
c2+a2/parenrightbig
+c/parenleftbig
a2+b2/parenrightbig
−/parenleftbig
a3+b3+c3/parenrightbig
2abc
=(a+b+c)(a2+b2+c2)−2(a3+b3+c3)
2abc
=4s/parenleftbig
s2−r2−4Rr/parenrightbig
−4/parenleftbig
s3−3r2s−6Rrs/parenrightbig
8Rrs
=/parenleftbig
s2−r2−4Rr/parenrightbig
−(s2−3r2−6Rr)
2Rr
=2r2+2Rr
2Rr=r
R+1.
/square
Example 2.5.2. IfA,BandCare the measurements of the angles in each of the
vertices of the triangle ABC, we have that cosA+c o s B+c o s C≤3
2.
Lemma 2.5.1 guarantees that cos A+cosB+cosC=r
R+1, and using Euler’s
inequality, R≥2r, we get the result.

72 Geometric Inequalities
We can give another direct proof. Observe that,
a(b2+c2−a2)+b(c2+a2−b2)+c(a2+b2−c2)=(b+c−a)(c+a−b)(a+b−c)+2abc.
Then,
cosA+c o s B+c o s C=b2+c2−a2
2bc+c2+a2−b2
2ca+a2+b2−c2
2ab
=(b+c−a)(c+a−b)(a+b−c)
2abc+1 ,
and since ( b+c−a)(c+a−b)(a+b−c)≤abc,w eh a v et h er e s u l t .
Example 2.5.3. (IMO, 1991) LetABC be a triangle, let Ibe its incenter and let
L,M,Nbe the intersections of the internal angle bisectors of A,B,CwithBC,
CA,AB, respectively. Prove that1
4<AI
ALBI
BMCI
CN≤8
27.
/Bullet
BCA
I
LM
Using the angle bisector theoremBL
LC=AB
CA=c
band the fact that BL+
LC=a, we can deduce that BL=ac
b+candLC=ab
b+c. Again, the angle bisector
theorem applied to the internal angle bisector BIof the angle ∠ABL gives us
IL
AI=BL
AB=ac
(b+c)c=a
b+c. Hence,
AL
AI=AI+IL
AI=1+IL
AI=1+a
b+c=a+b+c
b+c.
Then,AI
AL=b+c
a+b+c.13Similarly,BI
BM=c+a
a+b+candCI
CN=a+b
a+b+c. Therefore, the
inequality that we have to prove in terms of a,bandcis
1
4<(b+c)(c+a)(a+b)
(a+b+c)3≤8
27.
TheAM-GMinequality guarantees that
(b+c)(c+a)(a+b)≤/parenleftbigg(b+c)+(c+a)+(a+b)
3/parenrightbigg3
=8
27(a+b+c)3,
13Another way to prove the identity is as follows. Consider α=(ABI),β=(BCI)a n d
γ=(CAI). It is clear thatAI
AL=α+γ
α+β+γ=r(c+b)
r(a+c+b)=c+b
a+c+b.

2.5 Symmetric functions of a,bandc 73
hence the inequality on the right-hand side is now evident.
To prove the inequality on the left-hand side, ﬁrst note that
(b+c)(c+a)(a+b)
(a+b+c)3=(a+b+c)(ab+bc+ca)−abc
(a+b+c)3.
Substitute above, using equations (2.5), (2.6) and (2.7), to get
(b+c)(c+a)(a+b)
(a+b+c)3=2s(s2+r2+4Rr)−4Rrs
8s3
=2s3+2sr2+4Rrs
8s3=1
4+2r2+4Rr
8s2>1
4.
We can also use the Ravi transformation a=y+z,b=z+x,c=x+y,t or e a c h
the ﬁnal result in the following way:
(b+c)(c+a)(a+b)
(a+b+c)3=(x+y+z+x)(x+y+z+y)(x+y+z+z)
8(x+y+z)3
=1
8/parenleftbigg
1+x
x+y+z/parenrightbigg/parenleftbigg
1+y
x+y+z/parenrightbigg/parenleftbigg
1+z
x+y+z/parenrightbigg
=1
8/parenleftbigg
1+x+y+z
x+y+z+xy+yz+zx
x+y+z+xyz
x+y+z/parenrightbigg
>1
4.
Exercise 2.48. LetA,BandCbe the values of the angles in each one of the
vertices of the triangle ABC,p r o v et h a t
sin2A
2+s i n2B
2+s i n2C
2≥3
4.
Exercise 2.49. Leta,bandcbe the lengths of the sides of a triangle. Using the
tools we have studied in this section, prove that
4√
3(ABC)≤9abc
a+b+c.
Exercise 2.50. Leta,bandcbe the lengths of the sides of a triangle. Using the
tools presented in this section, prove that
4√
3(ABC)≤33√
a2b2c2.
Exercise 2.51. (IMO, 1961) Let a,bandcbe the lengths of the sides of a triangle,
prove that
4√
3(ABC)≤a2+b2+c2.
Exercise 2.52. Leta,bandcbe the lengths of the sides of a triangle, prove that
4√
3(ABC)≤a2+b2+c2−(a−b)2−(b−c)2−(c−a)2.

74 Geometric Inequalities
Exercise 2.53. Leta,bandcbe the lengths of the sides of a triangle, prove that
4√
3(ABC)≤ab+bc+ca.
Exercise 2.54. Leta,bandcbe the lengths of the sides of a triangle, prove that
4√
3(ABC)≤3(a+b+c)abc
ab+bc+ca.
Exercise 2.55. Leta,bandcbe the lengths of the sides of a triangle. If a+b+c=1 ,
prove that
a2+b2+c2+4abc <1
2.
Exercise 2.56. Leta,bandcbe the lengths of the sides of a triangle, let Randr
be the circumradius and the inradius, respectively, prove that
(b+c−a)(c+a−b)(a+b−c)
abc=2r
R.
Exercise 2.57. Leta,bandcbe the lengths of the sides of a triangle and let Rbe
the circumradius, prove that
3√
3R≤a2
b+c−a+b2
c+a−b+c2
a+b−c.
Exercise 2.58. Leta,bandcbe the lengths of the sides of a triangle. Set x=b+c−a
2,
y=c+a−b
2andz=a+b−c
2.I fτ1=x+y+z,τ2=xy+yz+zxandτ3=xyz,
verify the following relationships.
(1) (a −b)2+(b−c)2+(c−a)2=(x−y)2+(y−z)2+(z−x)2=2 (τ2
1−3τ2).
(2)a+b+c=2τ1.
(3)a2+b2+c2=2τ2
1−2τ2.
(4)ab+bc+ca=τ2
1+τ2.
(5)abc=τ1τ2−τ3.
(6) 16( ABC)2=2 (a2b2+b2c2+c2a2)−(a4+b4+c4)=1 6 r2s2=1 6τ1τ3.
(7)R=τ1τ2−τ3
4√
τ1τ3.
(8)r=/radicalbigg
τ3
τ1.
(9)τ1=s,τ2=r(4R+r),τ3=r2s.

2.6 Inequalities with areas and perimeters 75
2.6 Inequalities with areas and perimeters
We begin this section with the following example.
Example 2.6.1. (Austria–Poland, 1985) IfABCD is a convex quadrilateral of area
1,t h e n
AB+BC+CD+DA+AC+BD≥4+√
8.
Seta=AB,b=BC,c=CD,d=DA,e=ACandf=BD.The area of
the quadrilateral ABCD is (ABCD )=efsinθ
2,w h e r e θis the angle between the
diagonals, which makes it clear that 1 =efsinθ
2≤ef
2.
Since ( ABC)=absinB
2≤ab
2and (CDA)=cdsinD
2≤cd
2, we can deduce that
1=(ABCD )≤ab+cd
2.Similarly, 1 = ( ABCD )≤bc+da
2.These two inequalities
imply that ab+bc+cd+da≥4.
Finally, since ( e+f)2=4ef+(e−f)2≥4ef≥8a n d( a+b+c+d)2=
4(a+c)(b+d)+( (a+c)−(b+d))2≥4(a+c)(b+d)=4 ( ab+bc+cd+da)≥16,
we can deduce that a+b+c+d+e+f≥4+√
8.
Example 2.6.2. (Iberoamerican, 1992) Using the triangle ABC, construct a hexa-
gonHwith vertices A1,A2,B1,B2,C1,C2as shown in the ﬁgure. Show that the
area of the hexagon His at least thirteen times the area of the triangle ABC.
BCAA1A2
B1
B2C1C2aaa
b
b
bc
cc
It is clear, using the area formula ( ABC)=absinC
2,t h a t
(A1A2B1B2C1C2)= (A1BC 2)+(A2CB 1)+(B2AC 1)+(AA 1A2)
+(BB 1B2)+(CC 1C2)−2(ABC)
=(c+a)2sinB
2+(a+b)2sinC
2+(b+c)2sinA
2
+a2sinA
2+b2sinB
2+c2sinC
2−2(ABC)

76 Geometric Inequalities
=(a2+b2+c2)(sinA+s i nB+s i nC)
2+casinB
+absinC+bcsinA−2(ABC)
=(a2+b2+c2)(sinA+s i nB+s i nC)
2+4 (ABC).
Therefore, ( A1A2B1B2C1C2)≥13(ABC) if and only if
(a2+b2+c2)(sinA+s i nB+s i nC)
2≥9(ABC)=9abc
4R.
Using the sine law,sinA
a=1
2R, we can prove that the inequality is true if and only
if(a2+b2+c2)(a+b+c)
4R≥9abc
4R,t h a ti s ,
(a2+b2+c2)(a+b+c)≥9abc.
The last inequality can be deduced from the AM-GMinequality, from the re-
arrangement inequality or by using Tchebychev’s inequality. Moreover, the equal-
ity holds only in the case a=b=c.
Example 2.6.3. (China, 1988 and 1993) Consider two concentric circles of radii R
andR1(R1>R)and a convex quadrilateral ABCD inscribed in the small circle.
The extensions of AB,BC,CDandDAintersect the large circle at C1,D1,A1
andB1, respectively. Show that
(i)perimeter of A1B1C1D1
perimeter of ABCD≥R1
R;
(ii)(A1B1C1D1)
(ABCD )≥/parenleftbiggR1
R/parenrightbigg2
.
A1
B1
C1D1A
B
CD
O

2.6 Inequalities with areas and perimeters 77
To prove (i), we use Ptolemy’s inequality (see Exercise 2.11) applied to the
quadrilaterals OAB 1C1,OBC 1D1,OCD 1A1andODA 1B1, which implies that
AC 1·R1≤B1C1·R+AB 1·R1,
BD 1·R1≤C1D1·R+BC 1·R1, (2.8)
CA 1·R1≤D1A1·R+CD 1·R1,
DB 1·R1≤A1B1·R+DA 1·R1.
Then, when we add these inequalities together and write AC 1,BD 1,CA 1and
DB 1, and express them as AB+BC 1,BC+CD 1,CD+DA 1andDA+AB 1,
respectively, we get
R1·perimeter( ABCD )+R1(BC 1+CD 1+DA 1+AB 1)
≤R·perimeter( A1B1C1D1)+R1(AB 1+BC 1+CD 1+DA 1).
Therefore,
perimeter( A1B1C1D1)
perimeter( ABCD )≥R1
R.
To prove (ii), we use the fact that ( ABCD )=adsinA+bcsinA
2=sinA
2(ad+bc)
and also that ( ABCD )=absinB+cdsinB
2=sinB
2(ab+cd), where A=∠DAB and
B=∠ABC.
A1
B1
C1D1A
B
CD
Oa
bcdx
yzw
Since ( AB 1C1)=x(a+y)sin (180◦−A)
2=x(a+y)sinA
2, we can produce the identity
(AB 1C1)
(ABCD )=x(a+y)
ad+bc. Similarly,(BC 1D1)
(ABCD )=y(b+z)
ab+cd,(CD 1A1)
(ABCD )=z(c+w)
ad+bc,(DA 1B1)
(ABCD )=
w(d+x)
ab+cd. Then,
(A1B1C1D1)
(ABCD )=1+x(a+y)+z(w+c)
ad+bc+y(b+z)+w(d+x)
ab+cd.
The power of a point in the larger circle with respect to the small circle is equal to
R2
1−R2. In particular, the power of A1,B1,C1andD1is the same. On the other
hand, we know that these powers are w(w+c),x(x+d),y(y+a)a n d z(z+b),
respectively.

78 Geometric Inequalities
Substituting this in the previous equation implies that the area ratio is
(A1B1C1D1)
(ABCD )=1+( R2
1−R2)/bracketleftbiggx
y(ad+bc)+z
w(ad+bc)+y
z(ab+cd)+w
x(ab+cd)/bracketrightbigg
.
Using the AM-GMinequality allows us to deduce that
(A1B1C1D1)
(ABCD )≥1+4(R2
1−R2)
/radicalbig
(ad+bc)(ab+cd).
Since 2/radicalbig
(ad+bc)(ab+cd)≤ad+bc+ab+cd=(a+c)(b+d)≤1
4(a+b+c+d)2≤
(4√
2R)2
4=8R2, the ﬁrst two inequalities follow from the AM-GMinequality, and
the last one follows from the fact that, of all the quadrilaterals inscribed in a circle,
the square has the largest perimeter. Thus
(A1B1C1D1)
(ABCD )≥1+4(R2
1−R2)
4R2=/parenleftbiggR1
R/parenrightbigg2
.
Moreover, the equalities hold when ABCD is a square and only in this case. Since
in order to reduce inequalities (2.8) to identities, it must be the case that the fourquadrilaterals OAB
1C1,OBC 1D1,OCD 1A1andODA 1B1are cyclic. Thus, OA
is an internal angle bisector of the angle BAD,a n dt h es a m eh a p p e n sf o r OB,
OCandOD.
There are problems that, even when they are not presented in a geomet-
ric form, they invite us to search for geometric relationships, as in the followingexample.
Example 2.6.4. Ifa,b,care positive numbers with c<aandc<b, we can deduce
that/radicalbig
c(a−c)+/radicalbig
c(b−c)≤√
ab.
Consider the isosceles triangles ABC andACD, both sharing the common
sideACof length 2√
c; we take the ﬁrst triangle as having equal sides AB=BC
of length√
aand the second one satisfying CD=DAwith length√
b.
The area of the quadrilateral ABCD is, on the one hand,
(ABCD )=(ABC)+(ACD)=/radicalbig
c(a−c)+/radicalbig
b(b−c);
and on the other hand, ( ABCD )=2 ( ABD)=2√
absin∠BAD
2.
This last procedure for calculating the area clearly proves that ( ABCD )≤√
ab, and thus the result is obtained.

2.6 Inequalities with areas and perimeters 79
A
BCD
E
√
a√
a√
b√
b
√
c√
c
Another solution is as follows. Since ACandBDare perpendiculars, Py-
thagoras theorem implies that DE=√
b−candEB=√
a−c. By Ptolemy’s
inequality (see Exercise 2.11), (√
b−c+√
a−c)·(2√
c)≤√
a√
b+√
a√
band
then the result.
Exercise 2.59. On every side of a square with sides measuring 1, choose one point.
The four points will form a quadrilateral of sides of length a,b,candd,p r o v et h a t
(i) 2≤a2+b2+c2+d2≤4,
(ii) 2√
2≤a+b+c+d≤4.
Exercise 2.60. On each side of a regular hexagon with sides measuring 1, we choose
one point. The six points form a hexagon of perimeter h.P r o v et h a t3√
3≤h≤6.
Exercise 2.61. Consider the three lines tangent to the incircle of a triangle ABC
which are parallel to the sides of the triangle; these, together with the sides of the
triangle, form a hexagon T.P r o v et h a t
the perimeter of T≤2
3the perimeter of ( ABC).
Exercise 2.62. Find the radius of the circle of maximum area that can be covered
using three circles with radius 1.
Exercise 2.63. Find the radius of the circle of maximum area that can be covered
using three circles with radii r1,r2andr3.
Exercise 2.64. Two disjoint squares are located inside a square of side 1. If the
lengths of the sides of the two squares are aandb,p r o v et h a t a+b≤1.
Exercise 2.65. A convex quadrilateral is inscribed in a circumference of radius 1,
in such a way that one of its sides is a diameter and the other sides have lengths
a,bandc.P r o v et h a t abc≤1.

80 Geometric Inequalities
Exercise 2.66. LetABCDE be a convex pentagon such that the areas of the
triangles ABC,BCD,CDE,DEA andEAB are equal. Prove that
(i)(ABCDE )
4<(ABC)<(ABCDE )
3,
(ii) (ABCDE )=5+√
5
2(ABC).
Exercise 2.67. IfAD,BEandCFare the altitudes of the triangle ABC,p r o v e
that
perimeter( DEF)≤s,
where sis the semiperimeter.
Exercise 2.68. The lengths of the internal angle bisectors of a triangle are at most
1, show that the area of such a triangle is at most√
3
3.
Exercise 2.69. Ifa,b,c,dare the lengths of the sides of a convex quadrilateral,
show that
(i) (ABCD )≤ab+cd
2,
(ii) (ABCD )≤ac+bd
2and
(iii) (ABCD )≤/parenleftbigga+c
2/parenrightbigg/parenleftbiggb+d
2/parenrightbigg
.
2.7 Erd˝ os-Mordell Theorem
Theorem 2.7.1 (Pappus’s theorem). LetABCbe a triangle, AA/primeB/primeBandCC/primeA/prime/primeA
two parallelograms constructed on ACandABsuch that both either are inside
or outside the triangle. Let Pbe the intersection of B/primeA/primewithC/primeA/prime/prime.C o n s t r u c t
another parallelogram BP/primeP/prime/primeConBCsuch that BP/primeis parallel to APand of the
same length. Thus, we will have the following relationships between the areas:
(BP/primeP/prime/primeC)=(AA/primeB/primeB)+(CC/primeA/prime/primeA).
Proof. See the picture on the next page. /square

2.7 Erd˝ os-Mordell Theorem 81
P
AA/prime
A/prime/prime
BB/prime
CC/prime
P/primeP/prime/prime
Theorem 2.7.2 (Erd˝ os-Mordell theorem). LetPbe an arbitrary point inside or on
the boundary of the triangle ABC.I fpa,pb,pcare the distances from Pto the
sides of ABC, of lenghts a,b,c, respectively, then
PA+PB+PC≥2(pa+pb+pc).
Moreover, the equality holds if and only if the triangle ABC is equilateral and P
is the circumcenter.
Proof (Kazarinoﬀ) .Let us reﬂect the triangle ABC on the internal bisector BLof
angle B.L e tA/primeandC/primebe the reﬂections of AandC.T h ep o i n t Pis not reﬂected.
Now, let us consider the parallelograms determined by B,PandA/prime,a n db y B,P
andC/prime.
/BulletPA
C BC/prime
A/primeL
The sum of the areas of these parallelograms is cpa+apcand this is equal to the
area of the parallelogram A/primeP/primeP/prime/primeC/prime,w h e r e A/primeP/primeis parallel to BPand of the same
length. The area of A/primeP/primeP/prime/primeC/primeis at most b·PB. Moreover, the areas are equal if
BPis perpendicular to A/primeC/primeand this happens if and only if Pis onBO,w h e r e
Ois the circumcenter of ABC.14Then,
cpa+apc≤bPB.
14BPis perpendicular to A/primeC/primeif and only if ∠PBA/prime=9 0◦−∠A/prime, but∠A/prime=∠Aand
OBC =9 0◦−∠A,t h e n Pshould be on BO.

82 Geometric Inequalities
/BulletP
BC/prime
A/prime cpaa
pc /BulletPC/primeP/prime/prime
BP/prime
A/primeb
Therefore,
PB≥c
bpa+a
bpc.
Similarly,
PA≥b
apc+c
apband PC≥b
cpa+a
cpb.
If we add together these inequalities, we have
PA+PB+PC≥/parenleftbiggb
c+c
b/parenrightbigg
pa+/parenleftBigc
a+a
c/parenrightBig
pb+/parenleftbigga
b+b
a/parenrightbigg
pc≥2(pa+pb+pc),
sinceb
c+c
b≥2. Moreover, the equality holds if and only if a=b=candPis on
AO,BOandCO, that is, if the triangle is equilateral and P=O. /square
Example 2.7.3. Using the notation of the Erd˝ os-Mordell theorem, prove that
aPA+bPB+cPC≥4(ABC).
Consider the two parallelograms that are determined by B,C,PandB,
A,Pas shown in the ﬁgure, and the parallelogram that is constructed following
Pappus’s theorem. It is clear that
bPB≥apa+cpc.
/BulletP
BA
Cb c
apapc

2.7 Erd˝ os-Mordell Theorem 83
Similarly, it follows that
aPA≥bpb+cpc,
cPC≥apa+bpb.
Hence,
aPA+bPB+cPC≥2(apa+bpb+cpc)=4 ( ABC).
Example 2.7.4. Using the notation of the Erd˝ os-Mordell theorem, prove that
paPA+pbPB+pcPC≥2(papb+pbpc+pcpa).
As in the previous example, we have that aPA≥bpb+cpc.Hence,
paPA≥b
apapb+c
apcpa.
Similarly, we can deduce that pbPB≥a
bpapb+c
bpbpc,pcPC≥a
cpcpa+b
cpbpc.
If we add together these three inequalities, we get
paPA+pbPB+pcPC≥/parenleftbigga
b+b
a/parenrightbigg
papb+/parenleftbiggb
c+c
b/parenrightbigg
pbpc+/parenleftBigc
a+a
c/parenrightBig
pcpa
≥2(papb+pbpc+pcpa).
Example 2.7.5. Using the notation of the Erd˝ os-Mordell theorem, prove that
2/parenleftbigg1
PA+1
PB+1
PC/parenrightbigg
≤1
pa+1
pb+1
pc.
/BulletA
BCPA/prime
B/primeC/prime
A1B/prime
1
C1
A/prime
1C/prime
1
Let us apply inversion to the circle of center Pand radius d=pb.I fA/prime,B/prime,C/primeare
the inverse points of A,B,C, respectively, and A/prime
1,B/prime
1,C/prime
1are the inverse points

84 Geometric Inequalities
ofA1,B1,C1, we can deduce that
PA·PA/prime=PB·PB/prime=PC·PC/prime=d2,
PA 1·PA/prime
1=PB 1·PB/prime
1=PC 1·PC/prime
1=d2.
Moreover, A/prime,B/primeandC/primeare on B/prime
1C/prime
1,C/prime
1A/prime
1andA/prime1B/prime
1, respectively, and the
segments PA/prime,PB/primeandPC/primeare perpendicular to B/prime
1C/prime
1,C/prime
1A/prime
1andA/prime1B/prime
1, respec-
tively.
An application of the Erd˝ os-Mordell theorem to the triangle A/prime
1B/prime
1C/prime
1shows
thatPA/prime
1+PB/prime
1+PC/prime
1≥2(PA/prime+PB/prime+PC/prime).
Since
PA/prime
1=d2
PA 1,PB/prime
1=d2
PB 1,PC/prime
1=d2
PC 1,
PC/prime=d2
PC,PB/prime=d2
PB,PA/prime=d2
PA,
then
d2/parenleftbigg1
PA 1+1
PB 1+1
PC 1/parenrightbigg
≥2d2/parenleftbigg1
PA+1
PB+1
PC/parenrightbigg
,
that is,
2/parenleftbigg1
PA+1
PB+1
PC/parenrightbigg
≤/parenleftbigg1
pa+1
pb+1
pc/parenrightbigg
.
Example 2.7.6. Using the notation of the Erd˝ os-Mordell theorem, prove that
PA·PB·PC≥R
2r(pa+pb)(pb+pc)(pc+pa).
/BulletPA
C1 C Bbc
cpapc
LetC1be a point on BCsuch that BC 1=AB.Then AC 1=2csinB
2,a n d
Pappus’s theorem implies that PB/parenleftbig
2csinB
2/parenrightbig
≥cpa+cpc. Therefore,
PB≥pa+pc
2s i nB
2.

2.7 Erd˝ os-Mordell Theorem 85
Similarly,
PA≥pb+pc
2s i nA
2and PC≥pa+pb
2s i nC
2.
Then, after multiplication, we get
PA·PB·PC≥1
81
/parenleftbig
sinA
2/parenrightbig/parenleftbig
sinB
2/parenrightbig/parenleftbig
sinC
2/parenrightbig(pa+pb)(pb+pc)(pc+pa).
The solution of Exercise 2.42 helps us to prove that/parenleftbig
sinA
2/parenrightbig/parenleftbig
sinB
2/parenrightbig/parenleftbig
sinC
2/parenrightbig
=r
4R,
then the result follows.
Example 2.7.7. (IMO, 1991) LetPbe a point inside the triangle ABC. Prove that
at least one of the angles ∠PAB,∠PBC,∠PCAis less than or equal to 30◦.
Draw A1,B1andC1, the projections of Pon sides BC,CAandAB,r e –
spectively. Using the Erd˝ os-Mordell theorem we get PA+PB+PC≥2PA 1+
2PB 1+2PC 1.
A
BCA1B1 C1P
Thus, one of the following inequalities will be satisﬁed:
PA≥2PC 1,PB≥2PA 1orPC≥2PB 1.
If, for instance, PA≥2PC 1, we can deduce that1
2≥PC 1
PA=s i n∠PAB,t h e n
∠PAB≤30◦or∠PAB≥150◦.B u t ,i f ∠PAB≥150◦, then it must be the case
that∠PBC < 30◦a n dt h u si nb o t hc a s e st h er e s u l tf o l l o w s .
Example 2.7.8. (IMO, 1996) LetABCDEF be a convex hexagon such that AB
is parallel to DE,BCis parallel to EFandCDis parallel to FA.L e tRA,RC,
REdenote the circumradii of triangles FAB,BCD,DEF, respectively, and let
Pdenote the perimeter of the hexagon. Prove that
RA+RC+RE≥P
2.
LetM,NandPbe points inside the hexagon in such a way that MDEF ,
NFAB andPBCD are parallelograms. Let XYZ be the triangle formed by the

86 Geometric Inequalities
lines through B,D,Fand perpendicular to FA,BC,DE, respectively, where B
is on YZ,DonZXandFonXY.O b s e r v et h a t MNP andXYZ are similar
triangles.
X
YZA
BCDE
F
PMN
Since the triangles DEF andDMF are congruent, they have the same circumra-
dius; moreover, since XMis the diameter of the circumcircle of triangle DMF ,
thenXM=2RE. Similarly, YN=2RAandZP=2RC. Thus, the inequality
that needs to be proven can be written as
XM+YN+ZP≥BN+BP+DP+DM+FM+FN.
The case M=N=Pis the Erd˝ os-Mordell inequality, on which the rest of the
proof is based.
LetY/prime,Z/primedenote the reﬂections of YandZon the internal angle bisector
ofX.L e t G,Hdenote the feet of the perpendiculars of MandXonY/primeZ/prime,
respectively.
X
Z/primeY/primeD
HF
GM
Since ( XY/primeZ/prime)=(XMZ/prime)+(Z/primeMY/prime)+(Y/primeMX), we obtain
XH·Y/primeZ/prime=MF·XZ/prime+MG·Y/primeZ/prime+MD·Y/primeX.

2.7 Erd˝ os-Mordell Theorem 87
If we set x=Y/primeZ/prime,y=ZX/prime,z=XY/prime, the above equality becomes
xXH =xMG +zDM +yFM.
Since∠XHG =9 0◦,t h e n XH=XGsin∠XGH ≤XG. Moreover, using the
triangle inequality, XG≤XM+MG, we can deduce that
XM≥XH−MG=z
xDM+y
xFM.
Similarly,
YN≥x
yFN+z
yBN,
ZP≥y
zBP+x
yDP.
After adding together these three inequalities, we get
XM+YN+ZP≥z
xDM+y
xFM+x
yFN+z
yBN+y
zBP+x
zDP. (2.9)
Observe that
y
zBP+z
yBN=/parenleftbiggy
z+z
y/parenrightbigg/parenleftbiggBP+BN
2/parenrightbigg
+/parenleftbiggy
z−z
y/parenrightbigg/parenleftbiggBP−BN
2/parenrightbigg
.
Since the triangles XYZ andMNP are similar, we can deﬁne ras
r=FM−FN
XY=BN−BP
YZ=DP−DM
ZX.
If we apply the inequalityy
z+z
y≥2, we get
y
zBP+z
yBN=/parenleftbiggy
z+z
y/parenrightbigg/parenleftbiggBP+BN
2/parenrightbigg
−r
2/parenleftbiggyx
z−zx
y/parenrightbigg
≥BP+BN−r
2/parenleftbiggyx
z−zx
y/parenrightbigg
.
Similar inequalities hold for
x
yFN+y
xFM≥FN+FM−r
2/parenleftbiggxz
y−yz
x/parenrightbigg
,
z
xDM+x
zDP≥DM+DP−r
2/parenleftBigzy
x−xy
z/parenrightBig
.
If we add the inequalities and substitute them in (2.9), we have
XM+YN+ZP≥BN+BP+DP+DM+FM+FN,
which completes the proof.

88 Geometric Inequalities
Exercise 2.70. Using the notation of the Erd˝ os-Mordell theorem, prove that
PA·PB·PC≥4R
rpapbpc.
Exercise 2.71. Using the notation of the Erd˝ os-Mordell theorem, prove that
(i)PA2
pbpc+PB2
pcpa+PC2
papb≥12,
(ii)PA
pb+pc+PB
pc+pa+PC
pa+pb≥3,
(iii)PA
√
pbpc+PB
√
pcpa+PC
√
papb≥6,
(iv)PA·PB+PB·PC+PC·PA≥4(papb+pbpc+pcpa).
Exercise 2.72. LetABC be a triangle, Pbe an arbitrary point in the plane and
letpa,pbypcbe the distances from Pto the sides of a triangle of lengths a,band
c, respectively. If, for example, PandAare on diﬀerent sides of the segment BC,
thenpais negative, and we have a similar situation for the other cases. Prove that
PA+PB+PC≥/parenleftbiggb
c+c
b/parenrightbigg
pa+/parenleftBigc
a+a
c/parenrightBig
pb+/parenleftbigga
b+b
a/parenrightbigg
pc.
2.8 Optimization problems
In this section we present two classical examples known as the Fermat-Steiner
problem and the Fagnano problem.
The Fermat-Steiner problem . This problem seeks to ﬁnd a point in the interior or
on the sides of a triangle such that the sum of the distances from the point to the
vertices of the triangle is minimum. We will present three solutions and point out
the methods used to solve the problem.
Torricelli’s solution . It takes as its starting point the following two lemmas.
Lemma 2.8.1 (Viviani’s lemma). The sum of the distances from an interior point
to the sides of an equilateral triangle is equal to the altitude of the triangle.Proof. LetPbe a point in the interior of the triangle ABC. Draw the triangle
A
/primeB/primeC/primewith sides parallel to the sides of ABC,w i t h PonC/primeA/primeandB/primeC/primeon the
line through BandC.

2.8 Optimization problems 89
/BulletA
BCA/prime
B/primeC/primeL/primeMN
LPP/prime P/prime/primeM/prime
IfL,MandNare the feet of the perpendiculars of Pon the sides, it is clear that
PM=NM/prime,w h e r e M/primeis the intersection of PNwithA/primeB/prime.Moreover, PM/primeis
the altitude of the equilateral triangle AP/primeP.I fA/primeP/prime/primeis the altitude of the triangle
AP/primePfromA/prime, it is clear that PM/prime=A/primeP/prime/prime.L e tL/primebe a point on B/primeC/primesuch that
A/primeL/primeis the altitude of the triangle A/primeB/primeC/primefromA/prime.T h u s ,
PL+PM+PN=PL+PN+NM/prime=PL+A/primeP/prime/prime=A/primeP/prime/prime+P/prime/primeL/prime=A/primeL/prime./square
Next, we present another two proofs of Viviani’s lemma for the sake of com-
pleteness.
Observation 2.8.2. (i)The following is another proof of Viviani’s lemma which
is based on the use of areas. We have that (ABC)=(ABP)+(BCP)+
(CAP).T h e n ,i f ais the length of the side of the triangle and his the
length of its altitude, we have that ah=aPN+aPL+aPM, that is, h=
PN+PL+PM.
(ii)Another proof of Viviani’s lemma can be deduced from the following diagram.
/BulletA
BCMN
LP
M/prime

90 Geometric Inequalities
Lemma 2.8.3. IfABCis a triangle with all angles less than or equal to 120◦,t h e r e
is a unique point Psuch that ∠APB =∠BPC =∠CPA = 120◦.T h ep o i n t Pis
known as the Fermat point of the triangle.
Proof. First, we will proof the existence of P. On the sides ABandCAwe con-
struct equilateral triangles ABC/primeandCAB/prime. Their circumcircles intersect at A
and at another point that we denote as P.
B CA
C/primeB/prime
P
Since APCB/primeis cyclic, we have that ∠CPA = 180◦−∠B/prime= 120◦. Similarly, since
APBC/primeis cyclic, ∠APB = 120◦. Finally, ∠BPC = 360◦−∠APB−∠CPA =
360◦−120◦−120◦= 120◦.
To prove the uniqueness, suppose that Qsatisﬁes ∠AQB =∠BQC =
∠CQA = 120◦.S i n c e ∠AQB = 120◦,t h ep o i n t Qshould be on the circumcircle
ofABC/prime. Similarly, it should be on the circumcircle of CAB/prime, hence Q=P./square
We will now study Torricelli’s solution to the Fermat-Steiner problem. Given
the triangle ABC with angles less than or equal to 120◦, construct the Fermat
point P, which satisﬁes ∠APB =∠BPC =∠CPA = 120◦.N o w ,t h r o u g h A,B
andCwe draw perpendiculars to AP,BPandCP, respectively.
These perpendiculars determine a triangle DEF which is equilateral. This is
so because the quadrilateral PBDC is cyclic, having angles of 90◦inBandC.
Now, since ∠BPC = 120◦, we can deduce that ∠BDC =6 0◦. This argument can
be repeated for each angle. Therefore DEF is indeed equilateral.
We know that the distance from Pto the vertices of the triangle ABC is
equal to the length of the altitude of the equilateral triangle DEF.O b s e r v et h a t
any other point Qinside the triangle ABC satisﬁes AQ≥A/primeQ,w h e r e A/primeQis the
distance from Qto the side EF, similarly BQ≥B/primeQandCQ≥C/primeQ. Therefore
AQ+BQ+CQis greater than or equal to the altitude of DEF which is AP+
BP+CP, which in turn is equal to A/primeQ+B/primeQ+C/primeQas can be seen by using
Viviani’s lemma.

2.8 Optimization problems 91
BCA
P
DE
F
B/primeC/primeA/prime
Q
Hofmann-Gallai’s solution . This way of solving the problem uses the ingenious idea
of rotating the ﬁgure to place the three segments that we need next to each other,
in order to form a polygonal line and then add them together. Thus, when we join
the two extreme points with a segment of line, since this segment of line represents
the shortest path between them, it is then necessary to ﬁnd the conditions under
which the polygonal line lies over such segment. This proof was provided by J.Hofmann in 1929, but the method for proving had already been discovered and
should be attributed to the Hungarian Tibor Gallai. Let us recall this solution.
Consider the triangle ABC with a point Pinside it; draw AP,BPandCP.
Next, rotate the ﬁgure with its center in Band through an angle of 60
◦,i na
positive direction.
B CA
C/prime
P/prime
P60◦
We should point out several things. If C/primeis the image of AandP/primeis the image
ofP, the triangles BPP/primeandBAC/primeare equilateral. Moreover, if AP=P/primeC/prime
andBP=P/primeB=P/primeP,t h e n AP+BP+CP=P/primeC/prime+P/primeP+CP. The path
CP+PP/prime+P/primeCis minimum when C,P,P/primeandC/primeare collinear, which in turn
requires that ∠C/primeP/primeB= 120◦and∠BPC = 120◦; but since ∠C/primeP/primeB=∠APB,

92 Geometric Inequalities
the point Pshould satisfy ∠APB =∠BPC = 120◦(and then also ∠CPA = 120◦).
An advantage of this solution is that it provides another description of the
Fermat point and another way of ﬁnding it. If we review the proof, we can see that
the point Pis on the segment CC/prime,w h e r e C/primeis the third vertex of the equilateral
triangle with side AB. But if instead of the rotation with center in B, we rotate
it with its center in C, we obtain another equilateral triangle AB/primeCand we can
conclude that Pis onBB/prime. Hence we can ﬁnd Pas the intersection of BB/primeand
CC/prime.
Steiner’s solution. When we solve maximum and minimum problems we are prin-
cipally faced with three questions, (i) is there a solution?, (ii) is there a uniquesolution? (iii) what properties characterize the solution(s)? Torricelli’s solution
demonstrates that among all the points in the triangle, this particular point P,
from which the three sides of the triangle are observed as having an angle of 120
◦,
provides the minimum value of PA+PB+PC.I nt h i ss e n s e ,t h i sp o i n ta n s w e r s
the three questions we proposed and does so in an elegant way. However, the solu-
tion does not give us any clue as to why Torricelli chose this point, or what made
him choose that point; probably this will never be known. But in the following wecan consider a sequence of ideas that bring us to discover that the Fermat point
is the optimal point. These ideas belong to the Swiss geometer Jacob Steiner. Let
us ﬁrst provide the following two lemmas.
Lemma 2.8.4 (Heron’s problem). Given two points AandBon the same side of
al i n e d, ﬁnd the shortest path that begins at A, touches the line dand ﬁnishes at
B.
/Bullet/BulletB
PA
d
The shortest path between AandB, touching the line d, can be found re-
ﬂecting Bondto get a point B/prime;t h es e g m e n t AB/primeintersects dat a point P∗that
makes AP∗+P∗Brepresent the minimum between the numbers AP+PB,w i t h
Pond.
To convince ourselves it is suﬃcient to observe that
AP∗+P∗B=AP∗+P∗B/prime=AB/prime≤AP+PB/prime=AP+PB.
This point satisﬁes the following reﬂection principle: The incident angle is
equal to the reﬂection angle. It is evident that the point which has this property
is the minimum.

2.8 Optimization problems 93
/Bullet/Bullet
/BulletB
B/primeP P∗A
d α α
Lemma 2.8.5 (Heron’s problem using a circle). Given two points AandBoutside
the circle C, ﬁnd the shortest path that starts at A, touches the circle and ﬁnishes
atB.
/Bullet/Bullet
/BulletA
CB
We will only give a sketch of the solution.
LetDbe a point on C, then we have that the set {P:PA+PB=DA+DB}
is an ellipse EDwith foci points AandB, and that the point Dbelongs to ED.
In general Ed={P:PA+PB=d},w h e r e dis a positive number, is an ellipse
with foci AandB(ifd>A B ). Moreover, these ellipses have the property that Ed
is a subset of the interior of Ed/primeif and only if d<d/prime.
We would like to ﬁnd a point QonCsuch that the sum QA+QBis minimum.
The optimal point Qwill belong to an ellipse, precisely to EQ. Such an ellipse EQ
does not intersect Cin other point; in fact, if C/primeis another common point of EQ
andC,t h e ne v e r yp o i n t C/prime/primeof the circumference arc between QandC/primeofCwould
be in the interior of EQ, therefore C/prime/primeA+C/prime/primeB<Q A +QBand so Qis not the
optimal point, that is, a contradiction.
Thus, the point Qthat minimize AQ+QBshould satisfy that the ellipse EQ
is tangent to C. The common tangent line to EQandChappens to be perpendicular
to the radius CQ,w h e r e Cis the center of Cand, because of the reﬂection property
of the ellipse (the incidence angle is equal to the reﬂection angle), it follows that

94 Geometric Inequalities
the line CQis the internal bisector of the angle ∠AQB,t h a ti s , ∠BQC =∠CQA.
/Bullet
ABC
Qαβ
Now let us go back to Steiner’s solution of the Fermat-Steiner problem. A
point Pthat makes the sum PA+PB+PCa minimum can be one of the vertices
A,B,Cor a point of the triangle diﬀerent from the vertices. In the ﬁrst case, if
Pis one of the vertices, then one term of the sum PA+PB+PCis zero and the
other two are the lengths of the sides of the triangle ABC that have in common
the chosen vertex. Hence, the sum will be minimum when the chosen vertex is
opposite to the longest side of the triangle.
In order to analize the second case, Steiner follows the next idea (very useful
in optimization problems and one which can be taken to belong to the strategy of“divide and conquer” ), which is to keep ﬁxed some of the variables and optimize
the rest. This procedure would provide conditions in the variables not ﬁxed. Such
restrictions will act as restrains in the solution space until we reach the optimal
solution. Speciﬁcally, we proceed as follows. Suppose that PAis ﬁxed; that is, P
belongs to the circle of center Aand radius PA, where we need to ﬁnd the point
Pthat makes the sum PB+PCminimum. Note that Bshould be located outside
of such circle, otherwise PA≥ABand, using the triangle inequality, PB+PC >
BC. From this, it follows that PA+PB=PC >AB +BC,w h i c hm e a n s Bwould
be a more suitable point (instead of P). Similarly, Cshould be outside of such
c i r c l e .N o w ,s i n c e BandCare points outside the circle C=(A, PA), the optimal
point for the problem of minimizing PB+PCwith the condition that Pis on
the circle Cis, by Lemma 2.8.5, a point Qon the circle C, such that this circle is
tangent to the ellipse with foci BandCinQ,a n dt h ep o i n t Qis such that the
angles ∠AQB and∠CQA are equal. Since the role of A,B,Ccan be exchanged,
if now we ﬁx B(andPB), then the optimal point Qwill satisfy the condition
∠AQB =∠BQC and therefore ∠AQB =∠BQC =∠CQA = 120
◦. This means
Qshould be the Fermat point. All the above work in the second case is to assure
thatQis inside of ABC, if the angles of the triangle are not greater than 120◦.
The Fagnano problem. The problem is to ﬁnd an inscribed triangle of minimum
perimeter inside an acute triangle. We present two classical solutions, where thereﬂection on lines play a central role. One is due to H. Schwarz and the other toL. Fejer.

2.8 Optimization problems 95
Schwarz’s solution. The German mathematician Hermann Schwarz provided the
following solution to this problem for which he took as starting point two ob-
servations that we present as lemmas. These lemmas will demonstrate that theinscribed triangle with the minimum perimeter is the triangle formed using the
feet of the altitudes of the triangle. Such a triangle is known as the ortic triangle .
Lemma 2.8.6. LetABCbe a triangle, and let D,EandFbe the feet of the altitudes
onBC,CAandABas they fall from the vertices A,BandC, respectively. Then
the triangles ABC,AEF,DBF andDEC are similar.
Proof. It is suﬃcient to see that the ﬁrst two triangles are similar, since the other
cases are proved in a similar way.
/Bullet
BCA
E
F
H
D
Since these two triangles have a common angle at A, it is suﬃcient to see that
∠AEF =∠ABC.B u t ,s i n c ew ek n o wt h a t ∠AEF+∠FEC = 180◦and∠ABC+
∠FEC = 180◦because the quadrilateral BCEF is cyclic, then ∠AEF =∠ABC.
/square
Lemma 2.8.7. Using the notation of the previous lemma, we can deduce that the
reﬂection of DonABis collinear with EandF, and the reﬂection of DonCA
is collinear with EandF.
Proof. It follows directly from the previous lemma. /square
/Bullet
BCA
E
F
D/primeD/prime/prime
D

96 Geometric Inequalities
Using these elements we can now continue with the solution proposed by H.
Schwarz for the Fagnano problem.
We will now prove that the triangle with minimum perimeter is the ortic tri-
angle. Denote this triangle as DEF and consider another triangle LMN inscribed
inABC.
BC
AE
FDB
A
CA
BF/prime NN/primeM
L
Reﬂect the complete ﬁgure on the side BC, so that the resultant triangle is re-
ﬂected on CA,t h e no n AB,o nBCand ﬁnally on CA.
We have in total six congruent triangles and within each of them we have the
ortic triangle and the inscribed triangle LMN . The side ABof the last triangle
is parallel to the side ABof the ﬁrst, since as a result of the ﬁrst reﬂection, the
sideABis rotated in a negative direction through an angle of 2B ,a n dt h e ni na
negative direction through an angle of 2A , the third reﬂection is invariant and the
fourth is rotated through an angle of 2 Bin a positive direction and in the ﬁfth it
is also rotated in a positive direction through an angle of 2 A. Thus the total angle
of rotation of ABis zero.
The segment FF/primeis twice the perimeter of the ortic triangle, since FF/primeis
composed of six pieces where each side of the ortic triangle is taken twice. Also,
the broken line NN/primeis twice the perimeter of LMN. Moreover, NN/primeis parallel to
the line FF/primeand of the same length, then since the length of the broken line NN/prime
is greater than the length of the segment NN/prime, we can deduce that the perimeter
ofDEF is less than the perimeter of LMN.
The Fejer’s solution. The solution due to the Hungarian mathematician L.
Fejer also uses reﬂections. Let LMN be a triangle inscribed on ABC. Take both
the reﬂection L/primeof the point Lon the side CA,a n d L/prime/primethe reﬂection of Lon the
sideAB, and draw the segments ML/primeandNL/prime/prime. It is clear that LM=ML/primeand
L/prime/primeN=NL. Hence the perimeter of LMN satisﬁes
LM+MN+NL=L/prime/primeN+NM+ML/prime≥L/primeL/prime/prime.

2.8 Optimization problems 97
BCA
M
NL/prime
L/prime/prime
L
Thus, we can conclude that if we ﬁx the point L,t h ep o i n t s MandNthat make
the minimum perimeter LMN are the intersections of L/primeL/prime/primewithCAandAB,
respectively. Now, let us see which is the best option for the point L.W ea l r e a d y
know that the perimeter of LMN isL/primeL/prime/prime,t h u s Lshould make this quantity a
minimum.
BCA
M
NL/prime
L/prime/prime
L
It is evident that AL=AL/prime=AL/prime/primeand that ACandABare internal angle
bisectors of the angles LAL/primeandL/prime/primeAL, respectively. Thus ∠L/prime/primeAL/prime=2∠BAC =
2αwhich is a ﬁxed angle. The cosine law applied to the triangle AL/prime/primeL/primeguarantees
that
(L/primeL/prime/prime)2=(AL/prime)2+(AL/prime/prime)2−2AL/prime·AL/prime/primecos 2α
=2AL2(1−cos 2α).
Then, L/primeL/prime/primeis minimum when ALis minimum, which will be the case when ALis
the altitude.15A similar analysis using the points BandCwill demonstrate that
15This would be enough to ﬁnish Fejer’s proof for the Fagnano’s problem. This is the case
because if ALis the altitude and L/primeL/prime/primeintersects sides CAandABinEandF, respectively, then
BEandCFare altitudes. Let us see why. The triangle AL/prime/primeL/primeis isosceles with ∠L/prime/primeAL/prime=2∠A,
then∠AL/primeL/prime/prime=9 0◦−∠Aand by symmetry ∠ELA=9 0◦−∠A. Therefore ∠CLE =∠A.T h e n
AELB is a cyclic quadrilateral, therefore ∠AEB =∠ALB =9 0◦, which implies that BEis an
altitude. Similarly, it follows that CEis an altitude.

98 Geometric Inequalities
BMandCNshould also be altitudes. Thus, the triangle LMN with minimum
perimeter is the ortic triangle.
Exercise 2.73. LetABCD be a convex cyclic quadrilateral. If Ois the intersection
of the diagonals ACandBD,a n dP,Q,R,Sare the feet of the perpendiculars of O
on the sides AB,BC,CD,DA, respectively, prove that PQRS is the quadrilateral
of minimum perimeter inscribed in ABCD .
Exercise 2.74. LetPbe a point inside the triangle ABC.L e t D,EandFbe
the points of intersection of AP,BPandCPwith the sides BC,CAandAB,
respectively. Determine Psuch that the area of the triangle DEF is maximum.
Exercise 2.75. (IMO, 1981) Let Pbe a point inside the triangle ABC.L e tD,E,
Fbe the feet of the perpendiculars from Pto the lines BC,CA,AB, respectively.
Find the point Pthat minimizesBC
PD+CA
PE+AB
PF.
Exercise 2.76. LetP,D,EandFbe as in Exercise 2.75. For which point Pis
the sum of BD2+CE2+AF2minimum?
Exercise 2.77. LetP,D,EandFbe as in Exercise 2.75. For which point Pis
the product of PD·PE·PFmaximum?
Exercise 2.78. LetPbe a point inside the triangle ABC. For which point Pis the
sum of PA2+PB2+PC2minimum?
Exercise 2.79. For every point Pon the circumcircle of a triangle ABC,w ed r a w
the perpendiculars PMandPNto the sides ABandCA, respectively. Determine
for which point Pthe length MNis maximum and ﬁnd that length.
Exercise 2.80. (Turkey, 2000) Let ABC be an acute triangle with circumradius R;
letha,hbandhcbe the lengths of the altitudes AD,BEandCF, respectively.
Letta,tbandtcbe the lengths of the tangents from A,BandC, respectively, to
the circumcircle DEF.P r o v et h a t
t2
a
ha+t2
b
hb+t2
c
hc≤3
2R.
Exercise 2.81. Letha,hb,hcbe the lengths of the altitudes of a triangle ABC,
and let pa,pb,pcbe the distances from a point Pto the sides BC,CA,AB,
respectively, where Pis a point inside the triangle ABC.P r o v et h a t
(i)ha
pa+hb
pb+hc
pc≥9,
(ii)hahbhc≥27papbpc,
(iii) (ha−pa)(hb−pb)(hc−pc)≥8papbpc.
Exercise 2.82. Ifhis the length of the largest altitude of an acute triangle, then
r+R≤h.

2.8 Optimization problems 99
Exercise 2.83. Of all triangles with a common base and the same perimeter, the
isosceles triangle has the largest area.
Exercise 2.84. Of all triangles with a given perimeter, the one with largest area is
the equilateral triangle.Exercise 2.85. Of all inscribed triangles on a given circle, the one with largest
perimeter is the equilateral triangle.Exercise 2.86. IfPis a point inside the triangle ABC,l=PA,m=PBand
n=PC,p r o v et h a t
(lm+mn+nl)(l+m+n)≥a
2l+b2m+c2n.
Exercise 2.87. (IMO, 1961) Let a,bandcbe the lengths of the sides of a triangle
and let ( ABC) be the area of that triangle, prove that
4√
3(ABC)≤a2+b2+c2.
Exercise 2.88. Let (ABC) be the area of a triangle ABC and let Fbe the Fermat
point of the triangle. Prove that
4√
3(ABC)≤(AF+BF+CF)2.
Exercise 2.89. LetPbe a point inside the triangle ABC,p r o v et h a t
PA+PB+PC≥6r.
Exercise 2.90. (The area of the pedal triangle). For a triangle ABC and a point
Pon the plane, we deﬁne the “pedal triangle” ofPwith respect to ABC as the
triangle A1B1C1where A1,B1,C1are the feet of the perpendiculars from Pto
BC,CA,AB, respectively. Prove that
(A1B1C1)=(R2−OP2)(ABC )
4R2,
where Ois the circumcenter. We can thus conclude that the pedal triangle of
maximum area is the medial triangle.

Chapter 3
Recent Inequality Problems
Problem 3.1. (Bulgaria, 1995) Let SA,SBandSCdenote the areas of the reg-
ular heptagons A1A2A3A4A5A6A7,B1B2B3B4B5B6B7andC1C2C3C4C5C6C7,
respectively. Suppose that A1A2=B1B3=C1C4,p r o v et h a t
1
2<SB+SC
SA<2−√
2.
Problem 3.2. (Czech and Slovak Republics, 1995) Let ABCD be a tetrahedron
such that
∠BAC +∠CAD +∠DAB =∠ABC +∠CBD +∠DBA = 180◦.
Prove that CD≥AB.
Problem 3.3. (Estonia, 1995) Let a,b,cbe the lengths of the sides of a triangle
and let α,β,γbe the angles opposite to the sides. Prove that if the inradius of
the triangle is r,t h e n
asinα+bsinβ+csinγ≥9r.
Problem 3.4. (France, 1995) Three circles with the same radius pass through a
common point. Let Sbe the set of points which are interior to at least two of the
circles. How should the three circles be placed so that the area of Sis minimized?
Problem 3.5. (Germany, 1995) Let ABC be a triangle and let DandEbe points
onBCandCA, respectively, such that DEpasses through the incenter of ABC.
IfS=a r e a ( CDE)a n dr is the inradius of ABC,p r o v et h a t S≥2r2.
Problem 3.6. (Ireland, 1995) Let A,X,Dbe points on a line with Xbetween A
andD.L e tBbe a point such that ∠ABX = 120◦and let Cbe a point between
BandX.P r o v et h a t2 AD≥√
3(AB+BC+CD).

102 Recent Inequality Problems
Problem 3.7. (Korea, 1995) A ﬁnite number of points on the plane have the prop-
erty that any three of them form a triangle with area at most 1. Prove that all
these points lie within the interior or on the sides of a triangle with area less thanor equal to 4.
Problem 3.8. (Poland, 1995) For a ﬁxed positive integer n, ﬁnd the minimum
value of the sum
x
1+x2
2
2+x33
3+···+xnn
n,
given that x1,x2,…,xnare positive numbers satisfying that the sum of their
reciprocals is n.
Problem 3.9. (IMO, 1995) Let ABCDEF be a convex hexagon with AB=BC=
CDandDE=EF=FAsuch that ∠BCD =∠EFA =π
3.L e t GandHbe
points in the interior of the hexagon such that ∠AGB =∠DHE =2π
3.P r o v et h a t
AG+GB+GH+DH+HE≥CF.
Problem 3.10. (Balkan, 1996) Let Obe the circumcenter and Gthe centroid of the
triangle ABC.L e tRandrbe the circumradius and the inradius of the triangle.
Prove that OG≤/radicalbig
R(R−2r).
Problem 3.11. (China, 1996) Suppose that x0=0 ,xi>0f o ri=1,2,…,n ,a n d/summationtextn
i=1xi=1 .P r o v et h a t
1≤n/summationdisplay
i=1xi
√
1+x0+···+xi−1√
xi+···+xn<π
2.
Problem 3.12. (Poland, 1996) Let n≥2a n d a1,a2,…,a n∈R+with/summationtextn
i=1ai=1 .
Prove that for x1,x2,…,x n∈R+,w i t h/summationtextn
i=1xi=1 ,w eh a v e
2/summationdisplay
i<jxixj≤n−2
n−1+n/summationdisplay
i=1aix2
i
1−ai.
Problem 3.13. (Romania, 1996) Let x1,x2,…,xn,xn+1be positive real numbers
withx1+x2+···+xn=xn+1.P r o v et h a t
n/summationdisplay
i=1/radicalbig
xi(xn+1−xi)≤/radicaltp/radicalvertex/radicalvertex/radicalbt
n/summationdisplay
i=1xn+1(xn+1−xi).
Problem 3.14. (St. Petersburg, 1996) Let Mbe the intersection of the diagonals
of a cyclic quadrilateral, let Nbe the intersection of the segments that join the
opposite midpoints and let Obe the circumcenter. Prove that OM≥ON.

Recent Inequality Problems 103
Problem 3.15. (Austria–Poland, 1996) If w,x,yandzare real numbers satisfying
w+x+y+z=0a n d w2+x2+y2+z2=1 ,p r o v et h a t
−1≤wx+xy+yz+zw≤0.
Problem 3.16. (Taiwan, 1997) Let a1,…,anbe positive numbers such that
ai−1+ai+1
aiis an integer for all i=1,…,n ,a0=an,an+1=a1andn≥3.
Prove that
2n≤an+a2
a1+a1+a3
a2+a2+a4
a3+···+an−1+a1
an≤3n.
Problem 3.17. (Taiwan, 1997) Let ABC be an acute triangle with circumcenter
Oand circumradius R.P r o v et h a ti f AOintersects the circumcircle of OBC atD,
BOintersects the circumcircle of OCA atEandCOintersects the circumcircle
ofOAB atF,t h e n OD·OE·OF≥8R3.
Problem 3.18. (APMO, 1997) Let ABC be a triangle. The internal bisector of the
angle in Ameets the segment BCatXand the circumcircle at Y.L e tla=AX
AY.
Deﬁne lbandlcin the same way. Prove that
la
sin2A+lb
sin2B+lc
sin2C≥3
with equality if and only if the triangle is equilateral.
Problem 3.19. (IMO, 1997) Let x1,…,xnbe real numbers satisfying
|x1+···+xn|=1a n d |xi|≤n+1
2for all i=1,…,n . Prove that there exists
a permutation y1,…,ynofx1,…,xnsuch that
|y1+2y2+···+nyn|≤n+1
2.
Problem 3.20. (Czech and Slovak Republics, 1998) Let a,b,cbe positive real
numbers. A triangle exists with sides of lengths a,bandcif and only if there exist
numbers x,yandzsuch that
y
z+z
y=a
x,z
x+x
z=b
y,x
y+y
x=c
z.
Problem 3.21. (Hungary, 1998) Let ABCDEF be a centrally symmetric hexagon
and let P,Q,Rbe points on the sides AB,CD,EF, respectively. Prove that the
area of the triangle PQR is at most one-half of the area of the hexagon.
Problem 3.22. (Iran, 1998) Let x1,x2,x3andx4be positive real numbers such
thatx1x2x3x4=1 .P r o v et h a t
x3
1+x32+x33+x34≥max/braceleftbigg
x1+x2+x3+x4,1
x1+1
x2+1
x3+1
x4/bracerightbigg
.

104 Recent Inequality Problems
Problem 3.23. (Iran, 1998) Let x,y,zbe numbers greater than 1 and such that
1
x+1
y+1
z=2 .P r o v et h a t
√
x+y+z≥√
x−1+/radicalbig
y−1+√
z−1.
Problem 3.24. (Mediterranean, 1998) Let ABCD be a square inscribed in a circle.
IfMis a point on the arc AB,p r o v et h a t
MC·MD≥3√
3MA·MB.
Problem 3.25. (Nordic, 1998) Let Pbe a point inside an equilateral triangle ABC
of length side a. If the lines AP,BPandCPintersect the sides BC,CAandAB
of the triangle at L,MandN, respectively, prove that
PL+PM+PN <a.
Problem 3.26. (Spain, 1998) A line that contains the centroid Gof the triangle
ABC intersects the side ABatPand the side CAatQ.P r o v et h a t
PB
PA·QC
QA≤1
4.
Problem 3.27. (Armenia, 1999) Let Obe the center of the circumcircle of the
acute triangle ABC. The lines CO,AOandBOintersect the circumcircles of
the triangles AOB,BOC andAOC, for the second time, at C1,A1andB1,
respectively. Prove that
AA 1
OA 1+BB 1
OB 1+CC 1
OC 1≥9
2.
Problem 3.28. (Balkan, 1999) Let ABC be an acute triangle and let L,M,Nbe
the feet of the perpendiculars from the centroid GofABC to the sides BC,CA,
AB, respectively. Prove that
4
27<(LMN )
(ABC)≤1
4.
Problem 3.29. (Belarus, 1999) Let a,b,cbe positive real numbers such that
a2+b2+c2=3 .P r o v et h a t
1
1+ab+1
1+bc+1
1+ca≥3
2.
Problem 3.30. (Czech and Slovak Republics, 1999) For arbitrary positive numbers
a,bandc,p r o v et h a t
a
b+2c+b
c+2a+c
a+2b≥1.

Recent Inequality Problems 105
Problem 3.31. (Ireland, 1999) Let a,b,c,dbe positive real numbers with a+b+
c+d=1 .P r o v et h a t
a2
a+b+b2
b+c+c2
c+d+d2
d+a≥1
2.
Problem 3.32. (Italy, 1999) Let DandEbe given points on the sides ABandCA
of the triangle ABC such that DEis parallel to BCandDEis tangent to the
incircle of ABC.P r o v et h a t
DE≤AB+BC+CA
8.
Problem 3.33. (Poland, 1999) Let Dbe a point on the side BCof the triangle ABC
such that AD > BC .T h ep o i n t EonCAis deﬁned by the equationAE
EC=BD
AD−BC.
Prove that AD > BE .
Problem 3.34. (Romania, 1999) Let a,b,cbe positive real numbers such that
ab+bc+ca≤3abc.P r o v et h a t a+b+c≤a3+b3+c3.
Problem 3.35. (Romania, 1999) Let x1,x2,…,xnbe positive real numbers such
thatx1x2···xn=1 .P r o v et h a t
1
n−1+x1+1
n−1+x2+···+1
n−1+xn≤1.
Problem 3.36. (Romania, 1999) Let n≥2 be a positive integer and x1,y1,x2,y2,
…,xn,ynbe positive real numbers such that x1+x2+···+xn≥x1y1+x2y2+
···+xnyn.P r o v et h a t
x1+x2+···+xn≤x1
y1+x2
y2+···+xn
yn.
Problem 3.37. (Russia, 1999) Let a,bandcbe positive real numbers with abc=1 .
Prove that if a+b+c≤1
a+1
b+1
c,t h e n an+bn+cn≤1
an+1
bn+1
cnfor every
positive integer n.
Problem 3.38. (Russia, 1999) Let {x}=x−[x] denote the fractional part of x.
Prove that for every natural number n,
n2/summationdisplay
j=1/braceleftBig/radicalbig
j/bracerightBig
≤n2−1
2.
Problem 3.39. (Russia, 1999) The positive real numbers xandysatisfy x2+y3≥
x3+y4.P r o v et h a t
x3+y3≤2.

106 Recent Inequality Problems
Problem 3.40. (St. Petersburg, 1999) Let x0>x 1>···>xnbe real numbers.
Prove that
x0+1
x0−x1+1
x1−x2+···+1
xn−1−xn≥xn+2n.
Problem 3.41. (Turkey, 1999) Prove that ( a+3b)(b+4c)(c+2a)≥60abc for all
real numbers 0 ≤a≤b≤c.
Problem 3.42. (United Kingdom, 1999) Three non-negative real numbers a,band
csatisfy a+b+c=1 .P r o v et h a t
7(ab+bc+ca)≤2+9abc.
Problem 3.43. (USA, 1999) Let ABCD be a convex cyclic quadrilateral. Prove
that
|AB−CD|+|AD−BC|≥2|AC−BD|.
Problem 3.44. (APMO, 1999) Let {an}be a sequence of real numbers satisfying
ai+j≤ai+ajfor all i,j=1,2,….P r o v et h a t
a1+a2
2+···+an
n≥anfor all n∈N.
Problem 3.45. (IMO, 1999) Let n≥2b eaﬁ x e di n t e g e r .
(a) Determine the smallest constant Csuch that
/summationdisplay
1≤i<j≤nxixj(x2
i+x2j)≤C⎛
⎝/summationdisplay
1≤i≤nxi⎞⎠4
for all nonnegative real numbers x1,…,x n.
(b) For this constant Cdetermine when the equality occurs.
Problem 3.46. (Czech and Slovak Republics, 2000) Prove that for all positive real
numbers aandb,
3/radicalbigg
a
b+3/radicalbigg
b
a≤3/radicalBigg
2(a+b)/parenleftbigg1
a+1
b/parenrightbigg
.
Problem 3.47. (Korea, 2000) The real numbers a,b,c,x,y,zsatisfy a≥b≥c>0
andx≥y≥z>0. Prove that
a2x2
(by+cz)(bz+cy)+b2y2
(cz+ax)(cx+az)+c2z2
(ax+by)(ay+bx)≥3
4.
Problem 3.48. (Mediterranean, 2000) Let P,Q,R,Sbe the midpoints of the sides
BC,CD,DA,AB, respectively, of the convex quadrilateral ABCD .P r o v et h a t
4(AP2+BQ2+CR2+DS2)≤5(AB2+BC2+CD2+DA2).

Recent Inequality Problems 107
Problem 3.49. (Austria–Poland, 2000) Let x,y,zbe non-negative real numbers
such that x+y+z=1 .P r o v et h a t
2≤(1−x2)2+( 1−y2)2+( 1−z2)2≤(1 +x)(1 + y)(1 +z).
Problem 3.50. (IMO, 2000) Let a,b,cbe positive real numbers with abc=1 .
Prove that /parenleftbigg
a−1+1
b/parenrightbigg/parenleftbigg
b−1+1
c/parenrightbigg/parenleftbigg
c−1+1
a/parenrightbigg
≤1.
Problem 3.51. (Balkan, 2001) Let a,b,cbe positive real numbers such that abc≤
a+b+c.P r o v et h a t
a2+b2+c2≥√
3abc.
Problem 3.52. (Brazil, 2001) Prove that ( a+b)(a+c)≥2/radicalbig
abc(a+b+c), for all
positive real numbers a,b,c.
Problem 3.53. (Poland, 2001) Prove that the inequality
n/summationdisplay
i=1ixi≤/parenleftbiggn
2/parenrightbigg
+n/summationdisplay
i=1xi
i
holds for every integer n≥2 and for all non-negative real numbers x1,x2,…,x n.
Problem 3.54. (Austria–Poland, 2001) Prove that
2<a+b
c+b+c
a+c+a
b−a3+b3+c3
abc≤3,
where a,b,care the lengths of the sides of a triangle.
Problem 3.55. (IMO, 2001) Prove that for a,bandcpositive real numbers we
havea
√
a2+8bc+b
√
b2+8ca+c
√
c2+8ab≥1.
Problem 3.56. (Short list IMO, 2001) Let x1,x2,…,xnbe real numbers, prove
thatx1
1+x2
1+x2
1+x21+x22+···+xn
1+x21+···+x2n<√
n.
Problem 3.57. (Austria, 2002) Let a,b,cbe real numbers such that there exist
α,β,γ∈{ −1,1}withαa+βb+γc= 0. Determine the smallest positive value of/parenleftBig
a3+b3+c3
abc/parenrightBig2
.
Problem 3.58. (Balkan, 2002) Prove that
2
b(a+b)+2
c(b+c)+2
a(c+a)≥27
(a+b+c)2
for positive real numbers a,b,c.

108 Recent Inequality Problems
Problem 3.59. (Canada, 2002) Prove that for all positive real numbers a,b,c,
a3
bc+b3
ca+c3
ab≥a+b+c,
and determine when equality occurs.
Problem 3.60. (Ireland, 2002) Prove that
x
1−x+y
1−y+z
1−z≥33√
xyz
1−3√
xyz
for positive real numbers x,y,zless than 1.
Problem 3.61. (Rioplatense, 2002) Let a,b,cbe positive real numbers. Prove that
/parenleftbigga
b+c+1
2/parenrightbigg/parenleftbiggb
c+a+1
2/parenrightbigg/parenleftbiggc
a+b+1
2/parenrightbigg
≥1.
Problem 3.62. (Rioplatense, 2002) Let a,b,cbe positive real numbers. Prove that
a+b
c2+b+c
a2+c+a
b2≥9
a+b+c+1
a+1
b+1
c.
Problem 3.63. (Russia, 2002) Prove that√
x+√
y+√
z≥xy+yz+zxforx,y,
zpositive real numbers such that x+y+z=3 .
Problem 3.64. (APMO, 2002) Let a,b,cbe positive real numbers satisfying1
a+
1
b+1
c=1 .P r o v et h a t
√
a+bc+√
b+ca+√
c+ab≥√
abc+√
a+√
b+√
c.
Problem 3.65. (Ireland, 2003) The lengths a,b,cof the sides of a triangle are such
thata+b+c=2 .P r o v et h a t
1≤ab+bc+ca−abc≤1+1
27.
Problem 3.66. (Romania, 2003) Prove that in any triangle ABC the following
inequality holds:
1
mbmc+1
mcma+1
mamb≤√
3
S,
where Sis the area of the triangle and ma,mb,mcare the lengths of the medians.
Problem 3.67. (Romania, 2003) Let a,b,c,dbe positive real numbers with abcd=
1. Prove that
1+ab
1+a+1+bc
1+b+1+cd
1+c+1+da
1+d≥4.

Recent Inequality Problems 109
Problem 3.68. (Romania, 2003) In a triangle ABC,l e tla,lb,lcbe the lengths of
the internal angle bisectors, and let sbe the semiperimeter. Prove that
la+lb+lc≤√
3s.
Problem 3.69. (Russia, 2003) Let a,b,cbe positive real numbers with
a+b+c=1 .P r o v et h a t
1
1−a+1
1−b+1
1−c≥2
1+a+2
1+b+2
1+c.
Problem 3.70. (APMO, 2003) Prove that
(an+bn)1
n+(bn+cn)1
n+(cn+an)1
n<1+21
n
2,
where n>1 is an integer and a,b,care the side-lengths of a triangle with unit
perimeter.
Problem 3.71. (IMO, 2003) Given n>2a n dr e a ln u m b e r s x1≤x2≤··· ≤ xn,
prove that⎛
⎝/summationdisplay
i,j|xi−xj|⎞⎠2
≤2
3(n2−1)/summationdisplay
i,j(xi−xj)2,
where equality holds if and only if x1,x2,…,xnform an arithmetic progression.
Problem 3.72. (Short list Iberoamerican, 2004) If the positive numbers x1,x2,…,
xnsatisfy x1+x2+···+xn=1 ,p r o v et h a t
x1
x2(x1+x2+x3)+x2
x3(x2+x3+x4)+···+xn
x1(xn+x1+x2)≥n2
3.
Problem 3.73. (Czech and Slovak Republics, 2004) Let P(x)=ax2+bx+cbe
a quadratic polynomial with non-negative real coeﬃcients. Prove that for any
positive number x,
P(x)P/parenleftbigg1
x/parenrightbigg
≥(P(1))2.
Problem 3.74. (Croatia, 2004) Prove that the inequality
a2
(a+b)(a+c)+b2
(b+c)(b+a)+c2
(c+a)(c+b)≥3
4
holds for all positive real numbers a,b,c.
Problem 3.75. (Estonia, 2004) Let a,b,cbe positive real numbers such that
a2+b2+c2=3 .P r o v et h a t
1
1+2ab+1
1+2bc+1
1+2ca≥1.

110 Recent Inequality Problems
Problem 3.76. (Iran, 2004) Let x,y,zbe real numbers such that xyz=−1, prove
that
x4+y4+z4+3 (x+y+z)≥x2
y+x2
z+y2
x+y2
z+z2
x+z2
y.
Problem 3.77. (Korea, 2004) Let Randrbe the circumradius and the inradius
of the acute triangle ABC, respectively. Suppose that ∠Ais the largest angle of
ABC.L e tMbe the midpoint of BCand let Xbe the intersection of the tangents
to the circumcircle of ABC atBandC.P r o v et h a t
r
R≥AM
AX.
Problem 3.78. (Moldova, 2004) Prove that for all real numbers a,b,c≥0, the
following inequality holds:
a3+b3+c3≥a2√
bc+b2√
ca+c2√
ab.
Problem 3.79. (Ukraine, 2004) Let x,y,zbe positive real numbers with x+y+z=
1. Prove that
√
xy+z+√
yz+x+√
zx+y≥1+√
xy+√
yz+√
zx.
Problem 3.80. (Ukraine, 2004) Let a,b,cbe positive real numbers such that
abc≥1. Prove that
a3+b3+c3≥ab+bc+ca.
Problem 3.81. (Romania, 2004) Find all positive real numbers a,b,cwhich satisfy
the inequalities
4(ab+bc+ca)−1≥a2+b2+c2≥3(a3+b3+c3).
Problem 3.82. (Romania, 2004) The real numbers a,b,csatisfy a2+b2+c2=3 .
Prove the inequality
|a|+|b|+|c|−abc≤4.
Problem 3.83. (Romania, 2004) Consider the triangle ABC and let Obe a point in
the interior of ABC. The straight lines OA,OB,OCmeet the sides of the triangle
atA1,B1,C1, respectively. Let R1,R2,R3be the radii of the circumcircles of
the triangles OBC,OCA,OAB, respectively, and let Rbe the radius of the
circumcircle of the triangle ABC.P r o v et h a t
OA 1
AA 1R1+OB 1
BB 1R2+OC 1
CC 1R3≥R.
Problem 3.84. (Romania, 2004) Let n≥2 be an integer and let a1,a2,…,anbe
real numbers. Prove that for any non-empty subset S⊂{1,2,…,n }, the following
inequality holds:/parenleftBigg/summationdisplay
i∈Sai/parenrightBigg2
≤/summationdisplay
1≤i≤j≤n(ai+···+aj)2.

Recent Inequality Problems 111
Problem 3.85. (APMO, 2004) For any positive real numbers a,b,c,p r o v et h a t
(a2+2 ) (b2+2 ) (c2+2 )≥9(ab+bc+ca).
Problem 3.86. (Short list IMO, 2004) Let a,bandcbe positive real numbers such
thatab+bc+ca=1 .P r o v et h a t
33/radicalbigg
1
abc+6 (a+b+c)≤3√
3
abc.
Problem 3.87. (IMO, 2004) Let n≥3 be an integer. Let t1,t2,…,tnbe positive
real numbers such that
n2+1>(t1+t2+···+tn)/parenleftbigg1
t1+1
t2+···+1
tn/parenrightbigg
.
Prove that ti,tj,tkare the side-lengths of a triangle for all i,j,kwith 1 ≤i<
j<k≤n.
Problem 3.88. (Japan, 2005) Let a,bandcbe positive real numbers such that
a+b+c=1 .P r o v et h a t
a3√
1+b−c+b3√
1+c−a+a3√
1+a−b≤1.
Problem 3.89. (Russia, 2005) Let x1,x2,…,x6be real numbers such that x2
1+
x2
2+···+x26=6a n d x1+x2+···+x6=0 .P r o v et h a t x1x2···x6≤1
2.
Problem 3.90. (United Kingdom, 2005) Let a,b,cbe positive real numbers. Prove
that/parenleftbigga
b+b
c+c
a/parenrightbigg2
≥(a+b+c)/parenleftbigg1
a+1
b+1
c/parenrightbigg
.
Problem 3.91. (APMO, 2005) Let a,bandcbe positive real numbers such that
abc=8 .P r o v et h a t
a2
/radicalbig
(1 +a3)(1 + b3)+b2
/radicalbig
(1 +b3)(1 + c3)+c2
/radicalbig
(1 +c3)(1 + a3)≥4
3.
Problem 3.92. (IMO, 2005) Let x,y,zbe positive real numbers such that xyz≥1.
Prove that
x5−x2
x5+y2+z2+y5−y2
y5+z2+x2+z5−z2
z5+x2+y2≥0.
Problem 3.93. (Balkan, 2006) Let a,b,cbe positive real numbers, prove that
1
a(b+1 )+1
b(c+1 )+1
c(a+1 )≥3
1+abc.

112 Recent Inequality Problems
Problem 3.94. (Estonia, 2006) Let Obe the circumcenter of the acute triangle
ABC and let A/prime,B/primeandC/primebe the circumcenter of the triangles BCO,CAO and
ABO, respectively. Prove that the area of the triangle ABC is less than or equal
to the area of the triangle A/primeB/primeC/prime.
Problem 3.95. (Lithuania, 2006) Let a,b,cbe positive real numbers, prove that
1
a2+bc+1
b2+ca+1
c2+ab≤1
2/parenleftbigg1
ab+1
bc+1
ca/parenrightbigg
.
Problem 3.96. (Turkey, 2006) Let a1,a2,…,anbe positive real numbers such
that
a1+a2+···+an=a2
1+a22+···+a2n=A.
Prove that/summationdisplay
i/negationslash=jai
aj≥(n−1)2A
A−1.
Problem 3.97. (Iberoamerican, 2006) Consider nreal numbers a1,a2,…,an,
not necessarily distinct. Let dbe the diﬀerence between the maximum and the
minimum value of the numbers and let s=/summationtext
i<j|ai−aj|.P r o v et h a t
(n−1)d≤s≤n2d
4,
and determine the conditions on the nnumbers that ensure the validity of the
equalities.
Problem 3.98. (IMO, 2006) Determine the least real number Msuch that the
inequality
/vextendsingle/vextendsingleab(a2−b2)+bc(b2−c2)+ca(c2−a2)/vextendsingle/vextendsingle≤M(a2+b2+c2)2
is satisﬁed for all real numbers a,b,c.
Problem 3.99. (Bulgaria, 2007) Find all positive integers nsuch that if a,b,care
non-negative real numbers with a+b+c=3 ,t h e n
abc(an+bn+cn)≤3.
Problem 3.100. (Bulgaria, 2007) If a,b,care positive real numbers, prove that
(a+1 ) (b+1 )2
33√
c2a2+1+(b+1 ) (c+1 )2
33√
a2b2+1+(c+1 ) (a+1 )2
33√
b2c2+1≥a+b+c+3.
Problem 3.101. (China, 2007) If a,b,care the lengths of the sides of a triangle
witha+b+c= 3, ﬁnd the minimum of
a2+b2+c2+4abc
3.

Recent Inequality Problems 113
Problem 3.102. (Greece, 2007) If a,b,care the lengths of the sides of a triangle,
prove that
(a+b−c)4
b(b+c−a)+(b+c−a)4
c(c+a−b)+(c+a−b)4
a(a+b−c)≥ab+bc+ca.
Problem 3.103. (Iran, 2007) If a,b,care three diﬀerent positive real numbers,
prove that/vextendsingle/vextendsingle/vextendsingle/vextendsinglea+b
a−b+b+c
b−c+c+a
c−a/vextendsingle/vextendsingle/vextendsingle/vextendsingle>1.
Problem 3.104. (Mediterranean, 2007) Let x,y,zbe real numbers such that
xy+yz+zx=1 .P r o v et h a t xz < 1
2. Is it possible to improve the bound1
2?
Problem 3.105. (Mediterranean, 2007) Let x>1 be a real number which is not
an integer. Prove that
/parenleftbiggx+{x}
[x]−[x]
x+{x}/parenrightbigg
+/parenleftbiggx+[x]
{x}−{x}
x+[x]/parenrightbigg
>9
2,
where [ x]a n d{x}represent the integer part and the fractional part of x, respec-
tively.
Problem 3.106. (Peru, 2007) Let a,b,cbe positive real numbers such that a+b+
c≥1
a+1
b+1
c.P r o v et h a t
a+b+c≥3
a+b+c+2
abc.
Problem 3.107. (Romania, 2007) Let a,b,cbe positive real numbers such that
1
a+b+1+1
b+c+1+1
c+a+1≥1.
Prove that
a+b+c≥ab+bc+ca.
Problem 3.108. (Romania, 2007) Let ABC be an acute triangle with AB=AC.
For every interior point PofABC, consider the circle with center Aand radius
AP;l e tMandNbe the intersections of the sides ABandACwith the circle,
respectively. Determine the position of Pin such a way that MN+BP+CPis
minimum.Problem 3.109. (Romania, 2007) The points M,N,Pon the sides BC,CA,AB,
respectively, are such that the triangle MNP is acute. Let xbe the length of
the shortest altitude in the triangle ABC and let Xbe the length of the largest
altitude in the triangle MNP .P r o v et h a t x≤2X.

114 Recent Inequality Problems
Problem 3.110. (APMO, 2007) Let x,y,zbe positive real numbers such that√
x+√
y+√
z=1 .P r o v et h a t
x2+yz
/radicalbig
2×2(y+z)+y2+zx
/radicalbig
2y2(z+x)+z2+xy
/radicalbig
2z2(x+y)≥1.
Problem 3.111. (Baltic, 2008) If the positive real numbers a,b,csatisfy a2+b2+
c2=3 ,p r o v et h a t
a2
2+b+c2+b2
2+c+a2+c2
2+a+b2≥(a+b+c)2
12.
Under which circumstances the equality holds?
Problem 3.112. (Canada, 2008) Let a,b,cbe positive real numbers for which
a+b+c=1 .P r o v et h a t
a−bc
a+bc+b−ca
b+ca+c−ab
c+ab≤3
2.
Problem 3.113. (Iran, 2008) Find the least real number Ksuch that for any
positive real numbers x,y,z, the following inequality holds:
x√
y+y√
z+z√
x≤K/radicalbig
(x+y)(y+z)(z+x).
Problem 3.114. (Ireland, 2008) If the positive real numbers a,b,c,dsatisfy a2+
b2+c2+d2=1 ,p r o v et h a t
a2b2cd+ab2c2d+abc2d2+a2bcd2+a2bc2d+ab2cd2≤3
32.
Problem 3.115. (Ireland, 2008) Let x,y,zbe positive real numbers such that
xyz≥1. Prove that
(a) 27 ≤(1 +x+y)2+( 1+ y+z)2+( 1+ z+x)2,
(b) (1 + x+y)2+( 1+ y+z)2+( 1+ z+x)2≤3(x+y+z)2.
The equalities hold if and only if x=y=z=1 .
Problem 3.116. (Romania, 2008) If a,b,care positive real numbers with ab+bc+
ca=3 ,p r o v et h a t
1
1+a2(b+c)+1
1+b2(c+a)+1
1+c2(a+b)≤1
abc.
Problem 3.117. (Romania, 2008) Determine the maximum value for the real num-
berkif
(a+b+c)/parenleftbigg1
a+b+1
b+c+1
c+a−k/parenrightbigg
≥k
for all real numbers a,b,c≥0a n dw i t h a+b+c=ab+bc+ca.

Recent Inequality Problems 115
Problem 3.118. (Serbia, 2008) Let a,b,cbe positive real numbers such that
a+b+c=1 .P r o v et h a t
a2+b2+c2+3abc≥4
9.
Problem 3.119. (Vietnam, 2008) Let x,y,zbe distinct non-negative real numbers.
Prove that1
(x−y)2+1
(y−z)2+1
(z−x)2≥4
xy+yz+zx.
When is the case that the equality holds?
Problem 3.120. (IMO, 2008)
(i) If x,y,zare three real numbers diﬀerent from 1 and such that xyz=1 ,
prove that
x2
(x−1)2+y2
(y−1)2+z2
(z−1)2≥1.
(ii) Prove that the equality holds for an inﬁnite number of x,y,z,a l lo ft h e m
being rational numbers.

Chapter 4
Solutions to Exercises and
Problems
In this chapter we present solutions or hints to the exercises and problems that
appear in this book. In Sections 1 and 2 we provide the solutions to the exercises
in Chapters 1 and 2, respectively, and in Section 3 the solutions to the problems in
Chapter 3. We recommend that the reader should consult this chapter only afterhaving tried to solve the exercises or the problems by himself.
4.1 Solutions to the exercises in Chapter 1
Solution 1.1. It follows from the deﬁnition of a0.Also use ( −a)(−b)=ab.
(ii) (−a)b>0.
(iii)a<b⇔b−a>0, now use property 1.1.2.
(iv) Use property 1.1.2.(v) If a<0, then −a>0.
(vi)a
1
a=1>0.
(vii) If a<0, then −a>0.
(viii) Use (vi) and property 1.1.3.
(ix) Prove that ac < bc and that bc < bd .
(x) Use property 1.1.3 with a−1>0a n d a>0.
(xi) Use property 1.1.3 with 1 −a>0a n d a>0.
Solution 1.3. (i)a2<b2⇔b2−a2=(b+a)(b−a)>0.
(ii) If b>0, then1
b>0, now use Example 1.1.4.
Solution 1.4. For (i), (ii) and (iii) use the deﬁnition, and for (iv) and (v) remember
that|a|2=a2.

118 Solutions to Exercises and Problems
Solution 1.5. (i)x≤|x|and−x≤|x|.
(ii) Consider |a|=|a−b+b|and|b|=|b−a+a|, and apply the triangle inequal-
ity.
(iii) (x2+xy+y2)(x−y)=x3−y3.
(iv) (x2−xy+y2)(x+y)=x3+y3.
Solution 1.6. Ifa,borcis zero, the equality follows. Then, we can assume |a|≥
|b|≥|c|>0. Dividing by |a|, the inequality is equivalent to
1+/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsinglec
a/vextendsingle/vextendsingle/vextendsingle−/vextendsingle/vextendsingle/vextendsingle/vextendsingle1+b
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
a+c
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle−/vextendsingle/vextendsingle/vextendsingle1+c
a/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingle1+b
a+c
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥0.
Since/vextendsingle/vextendsingle
b
a/vextendsingle/vextendsingle≤1a n d/vextendsingle/vextendsingle
c
a/vextendsingle/vextendsingle≤1, we can deduce that/vextendsingle/vextendsingle1+
b
a/vextendsingle/vextendsingle=1 + b
aand/vextendsingle/vextendsingle1+ c
a/vextendsingle/vextendsingle=1 + c
a.
Thus, it is suﬃcient to prove that
/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsinglec
a/vextendsingle/vextendsingle/vextendsingle−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb
a+c
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle−/parenleftbigg
1+b
a+c
a/parenrightbigg
+/vextendsingle/vextendsingle/vextendsingle/vextendsingle1+b
a+c
a/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥0.
Now, use the triangle inequality and Exercise 1.5.
Solution 1.7. (i) Use that 0 ≤b≤1a n d1+ a>0i no r d e rt os e et h a t
0≤b(1 +a)≤1+a⇒0≤b−a≤1−ab⇒0≤b−a
1−ab≤1.
(ii) The inequality on the left-hand side is clear. Since 1 + a≤1+b, it follows
that1
1+b≤1
1+a, and then prove that
a
1+b+b
1+a≤a
1+a+b
1+a=a+b
1+a≤1.
(iii) For the inequality on the left-hand side, use that ab2−ba2=ab(b−a)i st h e
product of non-negative real numbers. For the inequality on the right-hand side,
note that b≤1⇒b2≤b⇒−b≤−b2,a n dt h e n
ab2−ba2≤ab2−b2a2=b2(a−a2)≤a−a2=1
4−(1
2−a)2≤1
4.
Solution 1.8. Prove in general that x<√
2⇒1+1
1+x>√
2a n dt h a t x>√
2⇒
1+1
1+x<√
2.
Solution 1.9. ax+by≥ay+bx⇔(a−b)(x−y)≥0.
Solution 1.10. We can assume that x≥y. Then, use the previous exercise substi-
tuting with√
x2,/radicalbig
y2,1
√
yand1
√
x.
Solution 1.11. Observe that
(a−b)(c−d)+(a−c)(b−d)+(d−a)(b−c)=2 ( a−b)(c−d)=2 ( a−b)2≥0.

4.1 Solutions to the exercises in Chapter 1 119
Solution 1.12. It follows from
f(a,c,b,d )−f(a,b,c,d )=(a−c)2−(a−b)2+(b−d)2−(c−d)2
=(b−c)(2a−b−c)+(b−c)(b+c−2d)
=2 (b−c)(a−d)>0,
f(a,b,c,d )−f(a, b, d, c )=(b−c)2−(b−d)2+(d−a)2−(c−a)2
=(d−c)(2b−c−d)+(d−c)(c+d−2a)
=2 (d−c)(b−a)>0.
Solution 1.13. In order for the expressions in the inequality to be well deﬁned, it
is necessary that x≥−1
2andx/negationslash= 0. Multiply the numerator and the denominator
by (1 +√
1+2x)2. Perform some simpliﬁcations and show that 2√
2x+1<7;
then solve for x.
Solution 1.14. Since 4 n2<4n2+n<4n2+4n+ 1, we can deduce that 2 n<√
4n2+n<2n+ 1. Hence, its integer part is 2 nand then we have to prove that√
4n2+n<2n+1
4, this follows immediately after squaring both sides of the
inequality.
Solution 1.15. Since ( a3−b3)(a2−b2)≥0, we have that a5+b5≥a2b2(a+b),
then
ab
a5+b5+ab≤ab
a2b2(a+b)+ab=abc2
a2b2c2(a+b)+abc2=c
a+b+c.
Similarly,bc
b5+c5+bc≤a
a+b+candca
c5+a5+ca≤b
a+b+c. Hence,
ab
a5+b5+ab+bc
b5+c5+bc+ca
c5+a5+ca≤c
a+b+c+a
a+b+c+b
a+b+c,
butc
a+b+c+a
a+b+c+b
a+b+c=c+a+b
a+b+c=1 .
Solution 1.16. Consider p(x)=ax2+bx+c, using the hypothesis, p(1) = a+b+c
andp(−1) =a−b+care not negative. Since a>0, the minimum value of pis
attained at−b
2aand its value is4ac−b2
4a<0.Ifx1,x2are the roots of p,w ec a n
deduce thatb
a=−(x1+x2)a n dc
a=x1x2, thereforea+b+c
a=( 1−x1)(1−x2),
a−b+c
a=( 1+ x1)(1 +x2)a n da−c
a=1−x1x2.Observe that, (1 −x1)(1−x2)≥0,
(1 +x1)(1 + x2)≥0a n d1 −x1x2≥0i m p l yt h a t −1≤x1,x2≤1.
Solution 1.17. If the inequalities are true, then a,bandcare less than 1, and
a(1−b)b(1−c)c(1−a)>1
64. On the other hand, since x(1−x)≤1
4for 0≤x≤1,
thena(1−b)b(1−c)c(1−a)≤1
64.
Solution 1.18. Use the AM-GM inequality with a=1 ,b=x.

120 Solutions to Exercises and Problems
Solution 1.19. Use the AM-GM inequality with a=x,b=1
x.
Solution 1.20. Use the AM-GM inequality with a=x2,b=y2.
Solution 1.21. In the previous exercise add x2+y2to both sides.
Solution 1.22. Use the AM-GM inequality with a=x+y
x,b=x+y
yand also use
the AM-GM inequality for xandy. Or reduce this to Exercise 1.20.
Solution 1.23. Use the AM-GM inequality with axandb
x.
Solution 1.24. Use the AM-GM inequality witha
bandb
a.
Solution 1.25.a+b
2−√
ab=(√
a−√
b)2
2, simplify and ﬁnd the bounds using 0 <b≤
a.
Solution 1.26. x+y≥2√
xy.
Solution 1.27. x2+y2≥2xy.
Solution 1.28. xy+zx≥2x√
yz.
Solution 1.29. See Exercise 1.27.
Solution 1.30.1
x+1
y≥2
√
xy.
Solution 1.31.xy
z+yz
x≥2/radicalBig
xy2z
zx=2y.
Solution 1.32.×2+(y2+z2)
2≥x/radicalbig
y2+z2.
Solution 1.33. x4+y4+8= x4+y4+4+4 ≥44/radicalbig
x4y416 = 8xy .
Solution 1.34. (a+b+c+d)≥44√
abcd,/parenleftbig1
a+1
b+1
c+1
d/parenrightbig
≥44/radicalBig
1
abcd.
Solution 1.35.a
b+b
c+c
d+d
a≥44/radicalBig
a
bb
cc
dd
a=4 .
Solution 1.36. (x1+···+xn)≥nn√
x1···xn,/parenleftBig
1
x1+···+1
xn/parenrightBig
≥nn/radicalBig
1
x1···xn.
Solution 1.37.a1
b1+a2
b2+···+an
bn≥nn/radicalBig
a1···an
b1···bn=n.
Solution 1.38. an−1>n/parenleftBig
an+1
2−an−1
2/parenrightBig
⇔(a−1)/parenleftbig
an−1+···+1/parenrightbig
>n an−1
2(a−
1)⇔an−1+···+a+1
n>an−1
2, but1+a+···+an−1
n>n/radicalbig
a(n−1)n
2=an−1
2.
Solution 1.39. 1=/parenleftbig1+a
2/parenrightbig/parenleftbig1+b
2/parenrightbig/parenleftbig1+c
2/parenrightbig
≥√
a√
b√
c=√
abc.
Solution 1.40. Using the AM-GM inequality, we obtain
a3
b+b3
c+bc≥33/radicalbigg
a3
b·b3
c·bc=3ab.

4.1 Solutions to the exercises in Chapter 1 121
Similarly,b3
c+c3
a+ca≥3bcandc3
a+a3
b+ab≥3ca. Therefore, 2(a3
b+b3
c+c3
a)+
(ab+bc+ca)≥3(ab+bc+ca).
Second solution. The inequality can also be proved using Exercise 1.107.
Solution 1.41. Ifabc= 0, the result is clear. If abc > 0, then we have
ab
c+bc
a+ca
b=1
2/parenleftbigg
a/parenleftbiggb
c+c
b/parenrightbigg
+b/parenleftBigc
a+a
c/parenrightBig
+c/parenleftbigga
b+b
a/parenrightbigg/parenrightbigg
≥1
2(2a+2b+2c),
and the result is evident.
Solution 1.42. Apply the AM-GM inequality twice over, a2b+b2c+c2a≥3abc,
ab2+bc2+ca2≥3abc.
Solution 1.43.1+ab
1+a=abc +ab
1+a=ab/parenleftBig
1+c
1+a/parenrightBig
,
1+ab
1+a+1+bc
1+b+1+ca
1+c=ab/parenleftbigg1+c
1+a/parenrightbigg
+bc/parenleftbigg1+a
1+b/parenrightbigg
+ca/parenleftbigg1+b
1+c/parenrightbigg
≥33/radicalBig
(abc)2=3.
Solution 1.44./parenleftBig
1
a+b+1
b+c+1
c+a/parenrightBig
(a+b+c)≥9
2is equivalent to
/parenleftbigg1
a+b+1
b+c+1
c+a/parenrightbigg
(a+b+b+c+c+a)≥9,
which follows from Exercise 1.36. For the other inequality use1
a+1
b≥4
a+b.S e e
Exercise 1.22.Solution 1.45. Note that
n+H
n
n=(1 + 1) + (1 +1
2)+···+( 1+1
n)
n.
Now, apply the AM-GM inequality.Solution 1.46. Setting y
i=1
1+xi,t h e n xi=1
yi−1=1−yi
yi.O b s e r v et h a t y1+···+
yn= 1 implies that 1 −yi=/summationtext
j/negationslash=iyi,t h e n/summationtext
j/negationslash=iyi≥(n−1)/parenleftBig/producttext
j/negationslash=iyj/parenrightBig 1
n−1and
/productdisplay
ixi=/productdisplay
i/parenleftbigg1−yi
yi/parenrightbigg
=/producttext
i/parenleftBig/summationtext
j/negationslash=iyj/parenrightBig
/producttext
iyi≥(n−1)n/producttext
i/parenleftBig/producttext
j/negationslash=iyj/parenrightBig 1
n−1
/producttext
iyi=(n−1)n.

122 Solutions to Exercises and Problems
Solution 1.47. Deﬁne an+1=1−(a1+···+an)a n d xi=1−ai
aifori=1,…,n +1.
Apply Exercise 1.46 directly.
Solution 1.48./summationtextn
i=11
1+ai=1⇒/summationtextni=1a
i
1+ai=n−1.Observe that
n/summationdisplay
i=1√
ai−(n−1)n/summationdisplay
i=11
√
ai=n/summationdisplay
i=11
1+ain/summationdisplay
i=1√
ai−n/summationdisplay
i=1ai
1+ain/summationdisplay
i=11
√
ai
=/summationdisplay
i,jai−aj
(1 +aj)√
ai=/summationdisplay
i>j(√
ai√
aj−1)(√
ai−√
aj)2(√
ai+√
aj)
(1 +ai)(1 + aj)√
ai√
aj.
Since 1 ≥1
1+ai+1
1+aj=2+ai+aj
1+ai+aj+aiaj, we can deduce that aiaj≥1. Hence the
terms of the last sum are positive.
Solution 1.49. LetSa=/summationtextn
i=1a2
i
ai+biandSb=/summationtextn
i=1b2
i
ai+bi.T h e n
Sa−Sb=n/summationdisplay
i=1a2
i−b2i
ai+bi=n/summationdisplay
i=1ai−n/summationdisplay
i=1bi=0 ,
thusSa=Sb=S.Hence, we have
2S=n/summationdisplay
i=1a2i+b2i
ai+bi≥1
2n/summationdisplay
i=1(ai+bi)2
ai+bi=n/summationdisplay
i=1ai,
where the inequality follows after using Exercise 1.21.
Solution 1.50. Since the inequality is homogeneous16we can assume that abc=1 .
Setting x=a3,y=b3andz=c3, the inequality is equivalent to
1
x+y+1+1
y+z+1+1
z+x+1≤1.
LetA=x+y+1 ,B=y+z+1a n d C=z+x+1 ,t h e n
1
A+1
B+1
C≤1⇔(A−1)(B−1)(C−1)−(A+B+C)+1≥0
⇔(x+y)(y+z)(z+x)−2(x+y+z)≥2
⇔(x+y+z)(xy+yz+zx−2)≥3.
Now, use that
x+y+z
3≥(xzy)1
3andxy+yz+zx
3≥(xyz)2
3.
16A function f(a, b, . . . ) is homogeneous if f(ta, tb, . . . )=tf(a ,b ,… )f o re a c h t∈R. Then, an
inequality of the form f(a, b, . . . )≥0, in the case of a homogeneous function, is equivalent to
f(t a ,t b ,… )≥0 for any t>0.

4.1 Solutions to the exercises in Chapter 1 123
Second solution. Follow the ideas used in the solutio n of Exercise 1.15. Start with
the inequality ( a2−b2)(a−b)≥0t og u a r a n t e et h a t a3+b3+abc≥ab(a+b+c),
then1
a3+b3+abc≤c
abc(a+b+c).
Solution 1.51. Note that abc≤/parenleftbiga+b+c
3/parenrightbig3=1
27.
/parenleftbigg1
a+1/parenrightbigg/parenleftbigg1
b+1/parenrightbigg/parenleftbigg1
c+1/parenrightbigg
=1+1
a+1
b+1
c+1
ab+1
bc+1
ca+1
abc
≥1+3
3√
abc+3
3/radicalBig
(abc)2+1
abc
=/parenleftbigg
1+1
3√
abc/parenrightbigg3
≥43.
Solution 1.52. The inequality is equivalent to/parenleftbigb+c
a/parenrightbig/parenleftbiga+c
b/parenrightbig/parenleftbiga+b
c/parenrightbig
≥8. Now, we use
the AM-GM inequality for each term of the product and the inequality follows
immediately.
Solution 1.53. Notice that
a
(a+1 ) (b+1 )+b
(b+1 ) (c+1 )+c
(c+1 ) (a+1 )
=(a+1 ) (b+1 ) (c+1 )−2
(a+1 ) (b+1 ) (c+1 )=1−2
(a+1 ) (b+1 ) (c+1 )≥3
4
if and only if ( a+1)(b+1)(c+1)≥8, and this last inequality follows immediately
from the inequality/parenleftbiga+1
2/parenrightbig/parenleftbigb+1
2/parenrightbig/parenleftbigc+1
2/parenrightbig
≥√
a√
b√
c=1 .
Solution 1.54. Observe that this exercise is similar to Exercise 1.52.
Solution 1.55. Apply the inequality between the arithmetic mean and the harmonic
mean to get
2ab
a+b=2
1
a+1
b≤a+b
2.
We can conclude that equality holds when a=b=c.
Solution 1.56. F i r s tu s et h ef a c tt h a t( a+b)2≥4ab, and then take into account
thatn/summationdisplay
i=11
aibi≥4n/summationdisplay
i=11
(ai+bi)2.
Now, use Exercise 1.36 to prove that
n/summationdisplay
i=1(ai+bi)2n/summationdisplay
i=11
(ai+bi)2≥n2.

124 Solutions to Exercises and Problems
Solution 1.57. Using the AM-GM inequality leads to xy+yz≥2y√
xz. Adding
similar results we get 2( xy+yz+zx)≥2(x√
yz+y√
zx+z√
xy). Once again,
using AM-GM inequality, we get x2+x2+y2+z2≥4x√
yz. Adding similar results
once more, we obtain x2+y2+z2≥x√
yz+y√
zx+z√
xy. Now adding both
results, we reach the conclusion(x+y+z)2
3≥x√
yz+y√
zx+z√
xy.
Solution 1.58. Using the AM-GM inequality takes us to x4+y4≥2x2y2. Applying
AM-GM inequality once again shows that 2x2y2+z2≥√
8xyz. Or, directly we
have that
x4+y4+z2
2+z2
2≥44/radicalbigg
x4y4z4
4=√
8xyz.
Solution 1.59. Use the AM-GM inequality to obtain
x2
y−1+y2
x−1≥2xy
/radicalbig
(x−1)(y−1)≥8.
The last inequality follows fromx
√
x−1≥2, since ( x−2)2≥0.
Second solution. Leta=x−1a n d b=y−1, which are positive numbers,
then the inequality we need to prove is equivalent to(a+1)2
b+(b+1)2
a≥8. Now,
by the AM-GM inequality we have ( a+1 )2≥4aand (b+1 )2≥4b. Then,
(a+1)2
b+(b+1)2
a≥4/parenleftbiga
b+b
a/parenrightbig
≥8. The last inequality follows from Exercise 1.24.
Solution 1.60. Observe that ( a,b,c)a n d( a2,b2,c2) have the same order, then use
inequality (1.2).
Solution 1.61. By the previous exercise
a3+b3+c3≥a2b+b2c+c2a.
Observe that (1
a,1
b,1
c)a n d(1
a2,1
b2,1
c2) can be ordered in the same way. Then, use
inequality (1.2) to get
(ab)3+(bc)3+(ca)3=1
a3+1
b3+1
c3
≥1
a21
c+1
b21
a+1
c21
b
=b
a+c
b+a
c
=a2b+b2c+c2a.
Adding these two inequalities leads to the result.
Solution 1.62. Use inequality (1.2) with ( a1,a2,a3)=(b1,b2,b3)=/parenleftbiga
b,b
c,c
a/parenrightbig
and
(a/prime
1,a/prime2,a/prime3)=/parenleftbigb
c,c
a,a
b/parenrightbig
.

4.1 Solutions to the exercises in Chapter 1 125
Solution 1.63. Use inequality (1.2) with ( a1,a2,a3)=(b1,b2,b3)=/parenleftbig1
a,1
b,1
c/parenrightbig
and
(a/prime
1,a/prime2,a/prime3)=/parenleftbig1
b,1
c,1
a/parenrightbig
.
Solution 1.64. Assume that a≤b≤c, and consider ( a1,a2,a3)=(a,b,c), then
use the rearrangement inequality (1.2) twice over with ( a/prime1,a/prime2,a/prime3)=(b,c,a)a n d
(c, a, b), respectively. Note that we are also using
(b1,b2,b3)=/parenleftbigg1
b+c−a,1
c+a−b,1
a+b−c/parenrightbigg
.
Solution 1.65. Use the same idea as in the previous exercise, but with nvariables.
Solution 1.66. Turn to the previous exercise and the fact thats
s−a1=1+a1
s−a1.
Solution 1.67. Apply Exercise 1.65 to the sequence a1,…,an,a1,…,an.
Solution 1.68. Apply Example 1.4.11.
Solution 1.69. Note that 1 = ( a2+b2+c2)+2(ab+bc+ca), and use the previous
exercise as follows:
1
3=a+b+c
3≤/radicalbigg
a2+b2+c2
3.
Therefore1
3≤a2+b2+c2. Hence, 2( ab+bc+ca)≤2
3, and the result is evident.
Second solution. The inequality is equivalent to 3(ab +bc+ca)≤(a+b+c)2,
which can be simpliﬁed to ab+bc+ca≤a2+b2+c2.
Solution 1.70. LetG=n√
x1x2···xnbe the geometric mean of the given numbers
and (a1,a2,…,a n)=/parenleftbigx1
G,x1x2
G2,…,x1x2···xn
Gn/parenrightbig
.
Using Corollary 1.4.2, we can establish that
n≤a1
a2+a2
a3+···+an−1
an+an
a1=G
x2+G
x3+···+G
xn+G
x1,
thusn
1
x1+···+1
xn≤G.
Also, using Corollary 1.4.2,
n≤a1
an+a2
a1+···+an
an−1=x1
G+x2
G+···+xn
G,
then
G≤x1+x2+···+xn
n.
The equalities hold if and only if a1=a2=···=an, that is, if and only if
x1=x2=···=xn.

126 Solutions to Exercises and Problems
Solution 1.71. The inequality is equivalent to
an−1
1+an−1
2+···+an−1
n≥a1···an
a1+a1···an
a2+···+a1···an
an,
which can be veriﬁed using the rearrangement inequality several times over.
Solution 1.72. First note that/summationtextn
i=1ai
√
1−ai=/summationtextni=11
√
1−ai−/summationtextni=1√
1−ai.U s e
the AM-GM inequality to obtain
1
nn/summationdisplay
i=11
√
1−ai≥n/radicaltp/radicalvertex/radicalvertex/radicalbt
n/productdisplay
i=11
√
1−ai=/radicalBigg
1
n/radicalbig
/producttextn
i=1(1−ai)
≥/radicalBigg
1
1
n/summationtextni=1(1−ai)=/radicalbigg
n
n−1.
Moreover,the Cauchy-Schwarz inequality serves to show that
n/summationdisplay
i=1√
1−ai≤/radicaltp/radicalvertex/radicalvertex/radicalbt
n/summationdisplay
i=1(1−ai)√
n=/radicalbig
n(n−1) andn/summationdisplay
i=1√
ai≤√
n.
Solution 1.73. (i)√
4a+1<4a+1+1
2=2a+1.
(ii) Use the Cauchy-Schwarz inequality with u=(√
4a+1,√
4b+1,√
4c+1 )a n d
v=( 1,1,1).
Solution 1.74. Suppose that a≥b≥c≥d(the other cases are similar). Then, if
A=b+c+d,B=a+c+d,C=a+b+dandD=a+b+c, we can deduce
that1
A≥1
B≥1
C≥1
D. Apply the Tchebyshev inequality twice over to show that
a3
A+b3
B+c3
C+d3
D≥1
4(a3+b3+c3+d3)/parenleftbigg1
A+1
B+1
C+1
D/parenrightbigg
≥1
16(a2+b2+c2+d2)(a+b+c+d)/parenleftbigg1
A+1
B+1
C+1
D/parenrightbigg
=1
16(a2+b2+c2+d2)/parenleftbiggA+B+C+D
3/parenrightbigg/parenleftbigg1
A+1
B+1
C+1
D/parenrightbigg
.
Now, use the Cauchy-Schwarz inequality to derive the result
a2+b2+c2+d2≥ab+bc+cd+da=1
and the inequality ( A+B+C+D)(1
A+1
B+1
C+1
D)≥16.
Solution 1.75. Apply the rearrangement inequality to
(a1,a2,a3)=/parenleftBigg
3/radicalbigg
a
b,3/radicalbigg
b
c,3/radicalbigg
c
a/parenrightBigg
,(b1,b2,b3)=⎛
⎝3/radicalbigg
/parenleftBiga
b/parenrightBig2
,3/radicalBigg
/parenleftbiggb
c/parenrightbigg2
,3/radicalbigg
/parenleftBigc
a/parenrightBig2⎞⎠

4.1 Solutions to the exercises in Chapter 1 127
and the permutation ( a/prime
1,a/prime2,a/prime3)=/parenleftbigg
3/radicalBig
b
c,3/radicalbig
c
a,3/radicalbig
a
b/parenrightbigg
to derive
a
b+b
c+c
a≥3/radicalbigg
a2
bc+3/radicalbigg
b2
ca+3/radicalbigg
c2
ab.
Finally, use the fact that abc=1 .
Second solution. The AM-GM inequality and the fact that abc= 1 imply that
1
3/parenleftbigga
b+a
b+b
c/parenrightbigg
≥3/radicalbigg
a
ba
bb
c=3/radicalbigg
a2
bc=3/radicalbigg
a3
abc=a.
Similarly,
1
3/parenleftbiggb
c+b
c+c
a/parenrightbigg
≥band1
3/parenleftBigc
a+c
a+a
b/parenrightBig
≥c,
and the result follows.
Solution 1.76. Using the hypothesis, for all k,l e a d st o s−2xk>0. Turn to the
Cauchy-Schwarz inequality to show that
/parenleftBiggn/summationdisplay
k=1×2
k
s−2xk/parenrightBigg/parenleftBiggn/summationdisplay
k=1(s−2xk)/parenrightBigg
≥/parenleftBiggn/summationdisplay
k=1xk/parenrightBigg2
=s2.
But 0 </summationtextn
k=1(s−2xk)=ns−2s, therefore
n/summationdisplay
k=1×2
k
s−2xk≥s
n−2.
Solution 1.77. The function f(x)=/parenleftbig
x+1
x/parenrightbig2is convex in R+.
Solution 1.78. The function
f(a,b,c)=a
b+c+1+b
a+c+1+c
a+b+1+( 1−a)(1−b)(1−c)
is convex in each variable, therefore its maximum is attained at the endpoints.
Solution 1.79. Ifx= 0, then the inequality reduces to 1 +1
√
1+y2≤2, which is
true because y≥0. By symmetry, the inequality holds for y=0 .
Now, suppose that 0 <x≤1a n d0 <y≤1. Let u≥0a n d v≥0 such that
x=e−uandy=e−v, then the inequality becomes
1
√
1+e−2u+1
√
1+e−2v≤2
√
1+e−(u+v),

128 Solutions to Exercises and Problems
that is,
f(u)+f(v)
2≤f/parenleftbiggu+v
2/parenrightbigg
,
where f(x)=1
√
1+e−2x.S i n c e f/prime/prime(x)=1−2e2x
(1+e−2x)5/2e4x, the function is concave in
the interval [0 ,∞). Thus the previous inequality holds.
Solution 1.80. Findf/prime/prime(x).
Solution 1.81. Use log(sin x)o rt h ef a c tt h a t
sinAsinB=s i n/parenleftbiggA+B
2+A−B
2/parenrightbigg
sin/parenleftbiggA+B
2−A−B
2/parenrightbigg
.
Solution 1.82. (i) If 1+ nx≤0, the inequality is evident since (1+ x)n≥0. Suppose
that (1 + nx)>0. Apply AM-GM inequality to the numbers (1 ,1,…,1,1+nx)
with ( n−1) ones.
(ii) Let a1,…,anbe positive numbers and deﬁne, for each j=1,…n,σj=
a1+···+aj
j. Apply Bernoulli’s inequality to show that/parenleftBig
σj
σj−1/parenrightBigj
≥jσj
σj−1−(j−1),
which implies
σj
j≥σj
j−1/parenleftbigg
jσj
σj−1−(j−1)/parenrightbigg
=σj−1
j−1(jσj−(j−1)σj−1)=ajσj−1
j−1.
Then, σn
n≥anσn−1
n−1≥anan−1σn−2
n−2≥···≥ anan−1···a1.
Solution 1.83. Ifx≥y≥z,w eh a v e xn(x−y)(x−z)≥yn(x−y)(y−z)a n d
zn(z−x)(z−y)≥0.
Solution 1.84. Notice that x(x−z)2+y(y−z)2−(x−z)(y−z)(x+y−z)≥0i fa n d
only if x(x−z)(x−y)+y(y−z)(y−x)+z(x−z)(y−z)≥0. The inequality now
follows from Sch¨ ur’s inequality. Alternatively, we can see that the last expression
is symmetric in x,yandz, then we can assume x≥z≥y, and if we return to the
original inequality, it becomes clear that
x(x−z)2+y(y−z)2≥0≥(x−z)(y−z)(x+y−z).
Solution 1.85. The inequality is homogeneous, therefore we can assume that a+
b+c= 1. Now, the terms on the left-hand side are of the formx
(1−x)2and the
function f(x)=x
(1−x)2is convex, since f/prime/prime(x)=4+2x
(1−x)4>0. By Jensen’s inequality
it follows thata
(1−a)2+b
(1−b)2+c
(1−c)2≥3f/parenleftbiga+b+c
3/parenrightbig
=3f/parenleftbig1
3/parenrightbig
=/parenleftbig3
2/parenrightbig2.
Solution 1.86. Since ( a+b+c)2≥3(ab+bc+ca), we can deduce that 1+3
ab+bc+ca≥
1+9
(a+b+c)2. Thus, the inequality will hold if
1+9
(a+b+c)2≥6
(a+b+c).

4.1 Solutions to the exercises in Chapter 1 129
But this last inequality follows from/parenleftBig
1−3
a+b+c/parenrightBig2
≥0.
Now, if abc=1 ,c o n s i d e rx =1
a,y=1
bandz=1
c; it follows immediately
thatxyz= 1. Thus, the inequality is equivalent to
1+3
xy+yz+zx≥6
x+y+z
which is the ﬁrst part of this exercise.
Solution 1.87. We will use the convexity of the function f(x)=xrforr≥1( i t s
second derivative is r(r−1)xr−2). First suppose that r>s> 0. Then Jensen’s
inequality for the convex function f(x)=xr
sapplied to xs
1,…,xsngives
t1xr
1+···+tnxrn≥(t1xs1+···+tnxsn)r
s
and taking the1
r-th power of both sides gives the desired inequality.
Now suppose 0 >r>s .T h e n f(x)=xr
sis concave, so Jensen’s inequality
is reversed; however, taking1
r-th powers reverses the inequality again.
Finally, in the case r>0>s,f(x)=xr
sis again convex, and taking1
r-th
powers preserves the inequality.
Solution 1.88. (i) Apply H¨ older’s inequality to the numbers xc
1,…,xc
n,yc
1,…,
yc
nwitha/prime=a
candb/prime=b
c.
(ii) Proceed as in Example 1.5.9. The only extra fact that we need to prove is
xiyizi≤1
axa
i+1
byb
i+1
czc
i, but this follows from part (i) of that example.
Solution 1.89. By the symmetry of the variables in the inequality we can assume
thata≤b≤c. We have two cases, (i) b≤a+b+c
3and (ii) b≥a+b+c
3.
Case (i): b≤a+b+c
3.
It happens thata+b+c
3≤a+c
2≤c, and it is true thata+b+c
3≤b+c
2≤c. Then,
there exist λ,μ∈[0,1] such that
c+a
2=λc+( 1−λ)/parenleftbigga+b+c
3/parenrightbigg
andb+c
2=μc+( 1−μ)/parenleftbigga+b+c
3/parenrightbigg
.
Adding these equalities, we obtain
a+b+2c
2=(λ+μ)c+( 2−λ−μ)/parenleftbigga+b+c
3/parenrightbigg
=( 2−λ−μ)/parenleftbigga+b−2c
3/parenrightbigg
+2c.
Hence,
a+b−2c
2=( 2−λ−μ)/parenleftbigga+b−2c
3/parenrightbigg
,
therefore 2 −(λ+μ)=3
2and (λ+μ)=1
2.

130 Solutions to Exercises and Problems
Now, since fis a convex function, we have
f/parenleftbigga+b
2/parenrightbigg
≤1
2(f(a)+f(b))
f/parenleftbiggb+c
2/parenrightbigg
≤μf(c)+( 1 −μ)f/parenleftbigga+b+c
3/parenrightbigg
f/parenleftbiggc+a
2/parenrightbigg
≤λf(c)+( 1 −λ)f/parenleftbigga+b+c
3/parenrightbigg
thus, adding these inequalities we get
f/parenleftbigga+b
2/parenrightbigg
+f/parenleftbiggb+c
2/parenrightbigg
+f/parenleftbiggc+a
2/parenrightbigg
≤1
2(f(a)+f(b)+f(c))
+3
2f/parenleftbigga+b+c
3/parenrightbigg
.
Case (ii): b≥a+b+c
3.
It is similar to case (i), using the fact that a≤a+c
2≤a+b+c
3anda≤a+b
2≤a+b+c
3.
Solution 1.90. If any of a,borcis zero, the inequality is evident. Applying Popovi-
ciu’s inequality (see the previous exercise) to the function f:R→R+deﬁned by
f(x)=e x p ( 2 x), which is convex since f/prime/prime(x)=4e x p ( 2 x)>0, we obtain
exp(2x)+e x p ( 2 y)+e x p ( 2 z)+3e x p/parenleftbigg2(x+y+z)
3/parenrightbigg
≥2[ e x p ( x+y)+e x p ( y+z)+e x p ( z+x)]
=2[ e x p ( x)exp (y)+e x p ( y)exp (z)+e x p ( z)exp (x)].
Setting a=e x p ( x),b=e x p ( y),c=e x p ( z), the previous inequality can be rewrit-
ten as
a2+b2+c2+33√
a2b2c2≥2(ab+bc+ca).
For the second part apply the AM-GM inequality in the following way:
2abc+1= abc+abc+1≥33√
a2b2c2.
Solution 1.91. Apply Popoviciu’s inequality to the convex function f(x)=x+1
x.
We will get the inequality1
a+1
b+1
c+9
a+b+c≥4
b+c+4
c+a+4
a+b. Then multiply
both sides by ( a+b+c) to ﬁnish the proof.
Solution 1.92. Observe that by using (1.8), we obtain
x2+y2+z2−|x||y|−|y||z|−|z||x|=1
2(|x|−|y|)2+1
2(|y|−|z|)2+1
2(|z|−|x|)2,
which is clearly greater than or equal to zero. Hence
|xy+yz+zx|≤|x||y|+|y||z|+|z||x|≤x2+y2+z2.
Second solution. Apply Cauchy-Schwarz inequality to ( x, y, z )a n d( y,z,x ).

4.1 Solutions to the exercises in Chapter 1 131
Solution 1.93. The inequality is equivalent to ab+bc+ca≤a2+b2+c2,w h i c h
we know is true. See Exercise 1.27.
Solution 1.94. Observe that if a+b+c= 0, then it follows from (1.7) that
a3+b3+c3=3abc.S i n c e( x−y)+(y−z)+(z−x) = 0, we can derive the
following factorization:
(x−y)3+(y−z)3+(z−x)3=3 (x−y)(y−z)(z−x).
Solution 1.95. Assume, without loss of generality, that a≥b≥c. We need to
prove that
−a3+b3+c3+3abc≥0.
Since
−a3+b3+c3+3abc=(−a)3+b3+c3−3(−a)bc,
the latter expression factors into
1
2(−a+b+c)((a+b)2+(a+c)2+(b−c)2).
The conclusion now follows from the triangle inequality, b+c>a.
Solution 1.96. Letp=|(x−y)(y−z)(z−x)|. Using AM-GM inequality on the
right-hand side of identity (1.8), we get
x2+y2+z2−xy−yz−zx≥3
23/radicalbig
p2. (4.1)
Now, since |x−y|≤x+y,|y−z|≤y+z,|z−x|≤z+x, it follows that
2(x+y+z)≥|x−y|+|y−z|+|z−x|. (4.2)
Applying again the AM-GM inequality leads to
2(x+y+z)≥33√
p,
and the result follows from inequalities (4.1) and (4.2).
Solution 1.97. Using identity (1.7), the condition x3+y3+z3−3xyz=1c a nb e
factorized as
(x+y+z)(x2+y2+z2−xy−yz−zx)=1. (4.3)
LetA=x2+y2+z2andB=x+y+z.N o t i c et h a t B2−A=2 (xy+yz+zx).
By identity (1.8), we have that B>0. Equation (4.3) now becomes
B/parenleftbigg
A−B2−A
2/parenrightbigg
=1,
therefore 3 A=B2+2
B.S i n c e B>0, we may apply the AM-GM inequality to
obtain
3A=B2+2
B=B2+1
B+1
B≥3,
that is, A≥1. For instance, the minimum A= 1 is attained when ( x, y, z )=
(1,0,0).

132 Solutions to Exercises and Problems
Solution 1.98. Inequality (1.11) helps to establish
1
a+1
b+4
c+16
d≥( 1+1+2+4 )2
a+b+c+d=64
a+b+c+d.
Solution 1.99. Apply inequality (1.11) twice over to get
a4+b4=a4
1+b4
1≥(a2+b2)2
2≥((a+b)2
2)2
2=(a+b)4
8.
Solution 1.100. Express the left-hand side as
(√
2)2
x+y+(√
2)2
y+z+(√
2)2
z+x
and use inequality (1.11).
Solution 1.101. Express the left-hand side as
x2
axy+bzx+y2
ayz+bxy+z2
azx+byz,
and then use inequality (1.11) to get
x2
axy+bzx+y2
ayz+bxy+z2
azx+byz≥(x+y+z)2
(a+b)(xy+yz+zx)≥3
a+b,
where the last inequality follows from (1.8).
Solution 1.102. Rewrite the left-hand side as
a2
a+b+b2
b+c+c2
a+c+b2
a+b+c2
b+c+a2
a+c,
and then apply inequality (1.11).
Solution 1.103. (i) Express the left-hand side as
x2
x2+2xy+3zx+y2
y2+2yz+3xy+z2
z2+2zx+3yz
and apply inequality (1.11) to get
x
x+2y+3z+y
y+2z+3x+z
z+2x+3y≥(x+y+z)2
x2+y2+z2+5 (xy+yz+zx).
Now it suﬃces to prove that
(x+y+z)2
x2+y2+z2+5 (xy+yz+zx)≥1
2,

4.1 Solutions to the exercises in Chapter 1 133
but this is equivalent to x2+y2+z2≥xy+yz+zx.
(ii) Proceed as in part (i), expressing the left-hand side as
w2
xw+2yw+3zw+x2
xy+2xz+3xw+y2
yz+2yw+3xy+z2
zw+2xz+3yz,
then use inequality (1.11) to get
w
x+2y+3z+x
y+2z+3w+y
z+2w+3x+z
w+2x+3y
≥(w+x+y+z)2
4(wx+xy+yz+zw+wy+xz).
Then, the inequality we have to prove becomes
(w+x+y+z)2
4(wx+xy+yz+zw+wy+xz)≥2
3,
which is equivalent to 3( w2+x2+y2+z2)≥2(wx+xy+yz+zw+wy+xz). This
follows by using the AM-GM inequality six times under the form x2+y2≥2xy.
Solution 1.104. We again apply inequality (1.11) to get
x2
(x+y)(x+z)+y2
(y+z)(y+x)+z2
(z+x)(z+y)
≥(x+y+z)2
x2+y2+z2+3 (xy+yz+zx).
Also, the inequality
(x+y+z)2
x2+y2+z2+3 (xy+yz+zx)≥3
4
is equivalent to
x2+y2+z2≥xy+yz+zx.
Solution 1.105. We express the left-hand side as
a2
a(b+c)+b2
b(c+d)+c2
c(d+a)+d2
d(a+b)
and apply inequality (1.11) to get
a2
a(b+c)+b2
b(c+d)+c2
c(d+a)+d2
d(a+b)≥(a+b+c+d)2
a(b+2c+d)+b(c+d)+d(b+c).

134 Solutions to Exercises and Problems
On the other hand, observe that
(a+b+c+d)2
(ac+bd)+(ab+ac+ad+bc+bd+cd)
=a2+b2+c2+d2+2ab+2ac+2ad+2bc+2bd+2cd
(ac+bd)+(ab+ac+ad+bc+bd+cd).
To prove that this last expression is greater than 2 is equivalent to showing that
a2+c2≥2acandb2+d2≥2bd, which can be done using the AM-GM inequality.
Solution 1.106. We express the left-hand side as
a2
ab+ac+b2
bc+bd+c2
cd+ce+d2
de+ad+e2
ae+be
and apply inequality (1.11) to get
a2
ab+ac+b2
bc+bd+c2
cd+ce+d2
de+ad+e2
ae+be≥(a+b+c+d+e)2
/summationtextab.
Since
(a+b+c+d+e)2=/summationdisplay
a2+2/summationdisplay
ab,
we have to prove that
2/summationdisplay
a2+4/summationdisplay
ab≥5/summationdisplay
ab,
which is equivalent to
2/summationdisplay
a2≥/summationdisplay
ab.
The last inequality follows from/summationtexta2≥/summationtextab.
Solution 1.107. (i) Using Tchebyshev’s inequality with the collections ( a≥b≥c)
and (a2
x≥b2
y≥c2
z), we obtain
1
3/parenleftbigga3
x+b3
y+c3
z/parenrightbigg
≥a2
x+b2
y+c2
z
3·a+b+c
3,
then by (1.11), we can deduce that
a2
x+b2
y+c2
z≥(a+b+c)2
x+y+z.
Therefore
a3
x+b3
y+c3
z≥(a+b+c)2
x+y+z·a+b+c
3.

4.1 Solutions to the exercises in Chapter 1 135
(ii) By Exercise 1.88, we have
/parenleftbigga3
x+b3
y+c3
z/parenrightbigg1
3
( 1+1+1 )1
3(x+y+z)1
3≥a+b+c.
Raising to the cubic power both sides and then dividing both sides by 3( x+y+z)
we obtain the result.
Solution 1.108. Using inequality (1.11), we obtain
x2
1+x22+···+x2n
x1+x2+···+xn
=x2
1
x1+x2+···+xn+x22
x1+x2+···+xn+···+x2n
x1+x2+···+xn
≥(x1+x2+···+xn)2
n(x1+x2+···+xn)=x1+x2+···+xn
n.
Thus, it is enough to prove that
/parenleftbiggx1+x2+···+xn
n/parenrightbiggkn
t
≥x1·x2·····xn.
Since k=m a x {x1,x2,…,x n}≥min{x1,x2,…,x n}=t,w eh a v et h a tkn
t≥n
and sincex1+x2+···+xn
n≥1, because all the xiare positive integers, it is enough
to prove that/parenleftbiggx1+x2+···+xn
n/parenrightbiggn
≥x1·x2·····xn,
which is equivalent to the AM-GM inequality.
Because all the intermediate inequalities are valid as equalities when x1=
x2=···=xn, we conclude that equality happens when x1=x2=···=xn.
Solution 1.109. Using the substitution a=x
y,b=y
zandc=z
x, the inequality
takes the form
a3
a3+2+b3
b3+2+c3
c3+2≥1,
and with the extra condition, abc=1 .
In order to prove this last inequality the extra condition is used as follows:
a3
a3+2+b3
b3+2+c3
c3+2=a3
a3+2abc+b3
b3+2abc+c3
c3+2abc
=a2
a2+2bc+b2
b2+2ca+c2
c2+2ab
≥(a+b+c)2
a2+b2+c2+2bc+2ca+2ab=1.
The inequality above follows from inequality (1.11).

136 Solutions to Exercises and Problems
Solution 1.110. With the substitution x=a
b,y=b
c,z=c
a, the inequality takes
the forma
b+c+b
c+a+c
a+b≥3
2,
which is Nesbitt’s inequality (Example 1.4.8).
Solution 1.111. Use the substitution x1=a2
a1,x2=a3
a2,…,xn=a1
an.S i n c e
1
1+x1+x1x2=1
1+a2
a1+a2
a1a3
a2=a1
a1+a2+a3and similarly for the other terms on the
left-hand side of the inequality, the inequality we have to prove becomes
a1
a1+a2+a3+a2
a2+a3+a4+···+an
an+a1+a2>1.
But this inequality is easy to prove. It is enough to observe that for all i=1,…,n
we have
ai+ai+1+ai+2<a 1+a2+···+an.
Solution 1.112. Using the substitution x=1
a,y=1
b,z=1
c, the condition ab+
bc+ca=abcbecomes x+y+z= 1 and the inequality is equivalent to
x4+y4
x3+y3+y4+z4
y3+z3+z4+x4
z3+x3≥1=x+y+z.
Tchebyshev’s inequality can be used to prove that
x4+y4
2≥x3+y3
2x+y
2,
thus
x4+y4
x3+y3+y4+z4
y3+z3+z4+x4
z3+x3≥x+y
2+y+z
2+z+x
2.
Solution 1.113. The inequality on the right-hand side follows from inequality
(1.11). For the inequality on the left-hand side, the substitution x=bc
a,y=ca
b,
z=ab
ctransforms the inequality into
x+y+z
3≥/radicalbigg
yz+zx+xy
3.
Squaring both sides, we obtain 3( xy+yz+zx)≤(x+y+z)2, which is valid if
and only if ( xy+yz+zx)≤x2+y2+z2, something we already know.
Solution 1.114. Note that
a−2
a+1+b−2
b+1+c−2
c+1≤0⇔3−3/parenleftbigg1
a+1+1
b+1+1
c+1/parenrightbigg
≤0
⇔1≤1
a+1+1
b+1+1
c+1.

4.1 Solutions to the exercises in Chapter 1 137
Using the substitution a=2x
y,b=2y
z,c=2z
x,w eg e t
1
a+1+1
b+1+1
c+1=1
2x
y+1+1
2y
z+1+1
2z
x+1
=y
2x+y+z
2y+z+x
2z+x
=y2
2xy+y2+z2
2yz+z2+x2
2zx+x2
≥(x+y+z)2
2xy+y2+2yz+z2+2zx+x2=1.
The only inequality in the expression follows from inequality (1.11).
Solution 1.115. Observe that
[5,0,0] =2
6(a5+b5+c5)≥2
6(a3bc+b3ca+c3ab)=[ 3,1,1],
where Muirhead’s theorem has been used.Solution 1.116. Using Heron’s formula for the area of a triangle, we can rewrite
the inequality as
a
2+b2+c2≥4√
3/radicalbigg
(a+b+c)
2(a+b−c)
2(a+c−b)
2(b+c−a)
2.
This is equivalent to
(a2+b2+c2)2≥3[((a+b)2−c2)(c2−(b−a)2)]
=3 ( 2c2a2+2c2b2+2a2b2−(a4+b4+c4)),
that is, a4+b4+c4≥a2b2+b2c2+c2a2, which, in terms of Muirhead’s theorem,
is equivalent to proving [4 ,0,0]≥[2,2,0].
Second solution. Using the substitution
x=a+b−c,y=a−b+c,z=−a+b+c,
we obtain x+y+z=a+b+c; then, using Heron’s formula we get
4(ABC)=/radicalbig
(a+b+c)(xyz)≤/radicalbigg
(a+b+c)(x+y+z)3
27=(a+b+c)2
3√
3.
Now we only need to prove that ( a+b+c)2≤3(a2+b2+c2). This last inequality
follows from Muirhead’s theorem, since [1 ,1,0]≤[2,0,0].

138 Solutions to Exercises and Problems
Solution 1.117. Notice that
a
(a+b)(a+c)+b
(b+c)(b+a)+c
(c+a)(c+b)≤9
4(a+b+c)
⇔8(ab+bc+ca)(a+b+c)≤9(a+b)(b+c)(c+a)
⇔24abc +8/summationdisplay
(a2b+ab2)≤9/summationdisplay
(a2b+ab2)+1 8 abc
⇔6abc≤a2b+ab2+b2c+bc2+c2a+ca2
⇔[1,1,1]≤[2,1,0].
Solution 1.118. The inequality is equivalent to
a3+b3+c3≥ab(a+b−c)+bc(b+c−a)+ca(c+a−b).
Setting x=a+b−c,y=b+c−a,z=a+c−b,w eg e ta =z+x
2,b=x+y
2,
c=y+z
2. Then, the inequality we have to prove is
1
8((z+x)3+(x+y)3+(y+z)3)≥1
4((z+x)(x+y)x+(x+y)(y+z)y+(y+z)(z+x)z),
which is again equivalent to
3(x2y+y2x+···+z2x)≥2(x2y+···)+6xyz
or
x2y+y2x+y2z+z2y+z2x+x2z≥6xyz,
and applying Muirhead’s theorem we obtain the result when x,y,zare non-
negative. If one of them is negative (and it cannot be more than one at a time),
we will get
x2(y+z)+y2(z+x)+z2(x+y)=x22c+y22a+z22b≥0
but 6xyzis negative, which ends the proof.
Solution 1.119. Observe that
a3
b2−bc+c2+b3
c2−ca+a2+c3
a2−ab+b2≥a+b+c
is equivalent to the inequality
a3(b+c)
b3+c3+b3(c+a)
c3+a3+c3(a+b)
a3+b3≥a+b+c,
w h i c hi nt u r ni se q u i v a l e n tt o
a3(b+c)(a3+c3)(a3+b3)+b3(c+a)(b3+c3)(a3+b3)
+c3(a+b)(a3+c3)(b3+c3)
≥(a+b+c)(a3+b3)(b3+c3)(c3+a3).

4.1 Solutions to the exercises in Chapter 1 139
The last inequality can be written in the terminology of Muirhead’s theorem as
[9,1,0] + [6 ,4,0] + [6 ,3,1] + [4 ,3,3]≥/parenleftbigg1
2[1,0,0]/parenrightbigg/parenleftbigg
[6,3,0] +1
3[3,3,3]/parenrightbigg
=[ 7,3,0] + [6 ,4,0] + [6 ,3,1] + [4 ,3,3]
⇔[9,1,0]≥[7,3,0],
a direct result of Muirhead’s theorem.
Solution 1.120. Suppose that a≤b≤c,t h e n
1
(1 +b)( 1+ c)≤1
(1 +c)( 1+ a)≤1
(1 +a)( 1+ b).
Use Tchebyshev’s inequality to prove that
a3
(1 +b)(1+ c)+b3
(1 +c)(1+ a)+c3
(1 +a)( 1+ b)
≥1
3(a3+b3+c3)/parenleftbigg1
(1 +b)(1 +c)+1
(1 +a)(1 + c)+1
(1 +a)(1 +b)/parenrightbigg
=1
3(a3+b3+c3)3+(a+b+c)
(1 +a)(1 + b)(1 +c).
Finally, use the facts that1
3(a3+b3+c3)≥(a+b+c
3)3,a+b+c
3≥1a n d( 1+ a)(1 +
b)(1 +c)≤/parenleftbig3+a+b+c
3/parenrightbig3to see that
1
3(a3+b3+c3)3+(a+b+a)
(1 +a)(1 +b)(1 + c)≥/parenleftbigga+b+c
3/parenrightbigg36
(1 +a+b+c
3)3≥6
8.
For the last inequality, notice thata+b+c
3
1+a+b+c
3≥1
2.
Second solution. Multiplying by the common denominator and expanding both
sides, the desired inequality becomes
4(a4+b4+c4+a3+b3+c3)≥3(1 + a+b+c+ab+bc+ca+abc).
Since 4( a4+b4+c4+a3+b3+c3) = 4(3[4, 0,0]+ 3[3, 0,0]) and 3(1 + a+b+c+
ab+bc+ca+abc)=3 ( [ 0 ,0,0] + 3[1, 0,0] + 3[1, 1,0] + [1, 1,1]), the inequality is
equivalent to
4[4,0,0] + 4[3, 0,0]≥[0,0,0] + 3[1, 0,0] + 3[1, 1,0] + [1 ,1,1].
Now, note that
[4,0,0]≥/bracketleftbigg4
3,4
3,4
3/bracketrightbigg
=a4
3b4
3c4
3=1=[ 0 ,0,0],

140 Solutions to Exercises and Problems
where it has been used that abc=1 .A l s o ,
3[4,0,0]≥3[2,1,1] = 31
3(a2bc+b2ca+c2ab)=31
3(a+b+c)=3 [ 1 ,0,0]
and
3[3,0,0]≥3/bracketleftbigg4
3,4
3,1
3/bracketrightbigg
=31
3/parenleftBig
a4
3b4
3c1
3+b4
3c4
3a1
3+c4
3a4
3b1
3/parenrightBig
=31
3(ab+bc+ca)=3 [ 1 ,0,0].
Finally, [3 ,0,0]≥[1,1,1]. Adding these results, we get the desired inequality.
4.2 Solutions to the exercises in Chapter 2
Solution 2.1. (i) Draw a segment BCof length a, a circle with radius cand center
inB, and a circle with radius band center in C, under what circumstances do
they intersect?
(ii) It follows from (i).
(iii)a=x+y,b=y+z,c=z+x⇔x=a+c−b
2,y=a+b−c
2,z=b+c−a
2.
Solution 2.2. (i)c<a +b⇒c<a +b+2√
ab=(√
a+√
b)2⇒√
c<√
a+√
b.
(ii) With 2, 3 and 4 it is possible to construct a triangle but with 4, 9 and 16 it is
not possible to do so.
(iii)a<b<c ⇒a+b<a+c<b+c⇒1
b+c<1
c+a<1
a+b, then it is suﬃcient to
see that1
a+b<1
b+c+1
c+a, and it will be even easier to see that1
c<1
b+c+1
c+a.
Solution 2.3. Use the fact that if a,b,care the lengths of the sides of a triangle,
the angle that is opposed to the side cis either 90◦or acute or obtuse if c2is equal,
less or greater than a2+b2, respectively. Now, suppose that a≤b≤c≤d≤e
and that the segments ( a,b,c)a n d( c, d, e) do not form an acute triangle; since
c2≥a2+b2ande2≥c2+d2, we deduce that e2≥a2+b2+d2≥a2+b2+c2≥
a2+b2+a2+b2=(a+b)2+(a−b)2≥(a+b)2, hence a+b≤e,w h i c hi sa
contradiction.
Solution 2.4. Since∠A>∠BthenBC > CA . Using the triangle inequality we
obtain AB < BC +CA, and by the previous statement, AB < 2BC.
Solution 2.5. (i) Let Obe the intersection point of the diagonals ACandBD.
Apply the triangle inequality to the triangles ABO andCDO. Adding the inequal-
ities, we get AB+CD < AC +BD. On the other hand, by hypothesis we have
thatAB+BD < AC +CD. Adding these last two inequalities we get AB < AC .
(ii) Let DEbe parallel to BC,t h e n∠EDA < ∠BCD < ∠A; therefore DE >1
2AD
and hence1
2AD < DE < BC . Refer to the previous exercise.

4.2 Solutions to the exercises in Chapter 2 141
Solution 2.6. Eachdiis less than the sum of the lengths of two sides. Also, use the
fact that in a convex quadrilateral the sum of the lengths of two opposite sides is
less than the sum of the lengths of the diagonals.
Solution 2.7. Use the triangle inequality in the triangles ABA/primeandAA/primeCto prove
thatc<m a+1
2aandb<m a+1
2a.
Solution 2.8. Ifα,β,γare the angles of a triangle in A,BandC, respectively, and
ifα1=∠BAA/primeandα2=∠A/primeAC, then, using D2, β>α 1andγ>α 2. Therefore,
180◦=α+β+γ>α 1+α2+α=2α. Or, if we draw a circle with diameter BC,
Ashould lie outside the circle and then ∠BAC < 90◦.
Solution 2.9. Construct a parallelogram ABDC , with one diagonal BCand the
other ADwhich is equal to two times the length of AA/primeand use D2 on the triangle
ABD.
Solution 2.10. Complete a parallelogram as in the previous solution to prove that
ma<b+c
2. Similarly, mb<a+c
2andmc<a+b
2.T op r o v et h el e f th a n ds i d e
inequality, let A/prime,B/primeandC/primebe the midpoints of the sides BC,CAandAB,
respectively.
BCA
A/primeC/primeA/prime/prime B/prime
Extend the segment C/primeB/primeto a point A/prime/primesuch that C/primeA/prime/prime=BC. Apply the previous
result to the triangle AA/primeA/prime/primewith side-lengths ma,mbandmc.
Solution 2.11. Consider the quadrilateral ABCD and let Obe a point on the
exterior of the quadrilateral so that AOB is similar to ACD, and thus OAC and
BAD are also similar. If O,BandCare collinear, we have an equality, otherwise
we have an inequality.17
Solution 2.12. Seta=AB,b=BC,c=CD,d=DA,m=ACandn=BD.
LetRbe the radius of the circumcircle of ABCD .T h u sw eh a v e18
(ABCD )=(ABC)+(CDA)=m(ab+cd)
4R,
(ABCD )=(BCD)+(DAB)=n(bc+ad)
4R.
17See [6, page 136] or [1, page 128].
18See [6, page 97] or [9, page 13].

142 Solutions to Exercises and Problems
Therefore
m
n=bc+ad
ab+cd>1⇔bc+ad > ab +cd
⇔(d−b)(a−c)>0.
Solution 2.13. Apply to the triangle ABP a rotation of 60◦with center at A.
Under the rotation the point Bgoes to the point C,a n dl e t P/primebe the image of
P. The triangle PP/primeCh a sa ss i d e s PP/prime=PA,P/primeC=PBandPC, and then the
result.
A
BCPP/primeA
BC
PP/prime
Second solution. Apply Ptolemy’s inequality (see Exercise 2.11) to the quadrilat-
eralsABCP ,ABPC andAPBC ; after cancellation of common terms we obtain
thatPB <PC +PA,PA<PC +PBandPC <PA +PB, respectively, which
establish the existence of the triangle.
Third solution. For the case when Pis inside ABC.L e tP/primebe the point where AP
intersects the side BC. Next, use that AP < AP/prime<A B =BC < PB +PC.I na
similar way, the other inequalities PB <PC +PAandPC <PA +PBhold.
Solution 2.14. Seta=AB,b=BC,x=AC,y=BD. Remember that in
a paralelogram we have 2( a2+b2)=x2+y2. We can suppose, without loss of
generality, that a≤b. It is clear that 2b<x +y, therefore (2b )2<(x+y)2=
x2+y2+2xy=2 (a2+b2)+2xy. Simplifying, we get 2( b2−a2)<2xy.
Solution 2.15. (i) Extend the medians AA/prime,BB/primeandCC/primeuntil they intersect the
circumcircle at A1,B1andC1, respectively. Use the power of A/primeto establish that
A/primeA1=a2
4ma. Also, use the facts that ma+A/primeA1≤2Rand that the length of
the median satisﬁes m2
a=2(b2+c2)−a2
4,t h a ti s ,4 m2a+a2=2 (b2+c2). We have
analogous expressions for mbandmc.
(ii) Use Ptolemy’s inequality in the quadrilaterals AC/primeGB/prime,BA/primeGC/primeandCB/primeGA/prime,
where Gdenotes the centroid. For instance, from the ﬁrst quadrilateral we get
2
3maa
2≤b
2mc
3+c
2mb
3,t h e n2 maa2≤abm c+cam b.

4.2 Solutions to the exercises in Chapter 2 143
Solution 2.16. Using the formula 4 m2
b+b2=2 (c2+a2), we observe that m2b−m2
c=
3
4(c2−b2). Now, using the triangle inequality, prove that mb+mc<3
2(b+c). From
this you can deduce the left-hand side inequality.
The right-hand side inequality can be obtained from the ﬁrst when applied
to the triangle of sides19with lengths ma,mbandmc.
Solution 2.17. Leta,b,cbe the lengths of the sides of ABC.I fEandFare the
projections of Iaon the sides ABandCA, respectively, it is clear that if rais the
radius of the excircle, we have that ra=IaE=EA=AF=FIa=s,w h e r e sis
the semiperimeter of ABC. Also, if hais the altitude of the triangle ABC from
vertex A,t h e nAD
DI a=ha
ra.S i n c e aha=bc,w eh a v et h a t
AD
DIa=ha
ra=bc
as=/parenleftbiggabc
4R/parenrightbigg/parenleftbigg4Rr
a2/parenrightbigg/parenleftbigg1
rs/parenrightbigg
=4Rr
a2,
where randRare the inradius and the circumradius of ABC, respectively. Since
2R=aand 2r=b+c−a, thereforeAD
DI a=b+c−a
a=b+c
a−1. Then, it is enough to
prove thatb+c
a≤√
2 or, equivalently, that 2 bc≤a2, butbc=√
b2c2≤b2+c2
2=a2
2.
Solution 2.18. Simplifying, the ﬁrst inequality is equivalent to ab+bc+ca≤
a2+b2+c2, which follows from Exercise 1.27. For the second one, expand ( a+b+c)2
and use the triangle inequality to obtain a2<a(b+c).
Solution 2.19. Use the previous suggestion.
Solution 2.20. Expand and you will get the previous exercise.
Solution 2.21. The ﬁrst inequality is the Nesbitt’s inequality, Example 1.4.8. For
the second inequality use the fact that a+b>a+b+c
2,t h e nc
a+b<2c
a+b+c.
Solution 2.22. Observe that a2(b+c−a)+b2(c+a−b)+c2(a+b−c)−2abc=
(b+c−a)(c+a−b)(a+b−c), now see Example 2.2.3.
Solution 2.23. Observe that
a/parenleftbig
b2+c2−a2/parenrightbig
+b/parenleftbig
c2+a2−b2/parenrightbig
+c/parenleftbig
a2+b2−c2/parenrightbig
=a2(b+c−a)+b2(c+a−b)+c2(a+b−c),
now see Exercise 2.22.
Solution 2.24. Use Ravi’s transformation with a=y+z,b=z+x,c=x+yto
see ﬁrst that
a2b(a−b)+b2c(b−c)+c2a(c−a)=2 ( xy3+yz3+zx3)−2(xy2z+x2yz+xyz2).
Then, the inequality is equivalent tox2
y+y2
z+z2
x≥x+y+z. Apply then inequality
(1.11).
19See the solution of Exercise 2.10.

144 Solutions to Exercises and Problems
Solution 2.25.
/vextendsingle/vextendsingle/vextendsingle/vextendsinglea−b
a+b+b−c
b+c+c−a
c+a/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsinglea−b
a+b·b−c
b+c·c−a
c+a/vextendsingle/vextendsingle/vextendsingle/vextendsingle
<cab
(a+b)(b+c)(c+a)≤1
8.
For the last inequality, see the solution of Example 2.2.3.
Solution 2.26. By Exercise 2.18,
3(ab+bc+ca)≤(a+b+c)2≤4(ab+bc+ca).
Then, since ab+bc+ca= 3, it follows that 9 ≤(a+b+c)2≤12, and then the
result.Solution 2.27. Use Ravi’s transformation, a=y+z,b=z+xandc=x+y.T h e
AM-GM inequality and the Cauchy-Schwarz inequality imply
1
a+1
b+1
c=1
y+z+1
z+x+1
x+y
≤1
2/parenleftbigg1
√
yz+1
√
zx+1
√
xy/parenrightbigg
=√
x+√
y+√
z
2√
xyz
≤√
3√
x+y+z
2√
xyz
=√
3
2/radicalbigg
x+y+z
xyz=√
3
2r.
For the last identity, see the end of the proof of Example 2.2.4.Solution 2.28. The part (i) follows from the following equivalences:
(s−a)(s−b)<a b⇔s
2−s(a+b)<0
⇔a+b+c<2(a+b)
⇔c<a +b.
For (ii), use Ravi’s transformation, a=y+z,b=z+x,c=x+y,i no r d e r
to see that the inequality is equivalent to
4(xy+yz+zx)≤(y+z)(z+x)+(z+x)(x+y)+(x+y)(y+z).
In turn, the last inequality follows from the inequality xy+yz+zx≤x2+y2+z2,
which is Exercise 1.27.

4.2 Solutions to the exercises in Chapter 2 145
Another way to obtain (ii) is the following: the given inequality is equivalent
to 3s2−2s(a+b+c)+(ab+bc+ca)≤ab+bc+ca
4, which in turn is equivalent to
3(ab+bc+ca)≤4s2. The last inequality can be rewritten as 3( ab+bc+ca)≤
(a+b+c)2.
Solution 2.29. Applying the cosine law, we can see that
/radicalbig
a2+b2−c2/radicalbig
a2−b2+c2=√
2abcosC√
2accosB
=2a/radicalbig
(bcosC)(ccosB)
≤2abcosC+ccosB
2=a2.
Solution 2.30. Using the Cauchy-Schwarz inequality, for any x,y,z,w≥0, we
have that√
xy+√
zw≤/radicalbig
(x+z)(y+w).
Therefore
/summationdisplay
cyclic/radicalbig
a2+b2−c2/radicalbig
a2−b2+c2=1
2/summationdisplay
cyclic/parenleftBig/radicalbig
a2+b2−c2/radicalbig
a2−b2+c2
+/radicalbig
c2+a2−b2/radicalbig
c2−a2+b2/parenrightBig
≤1
2/summationdisplay
cyclic/radicalbig
(2a2)(2c2)=/summationdisplay
cyclicac.
Solution 2.31. Consider positive numbers x,y,zwitha=y+z,b=z+xand
c=x+y.The inequalities are equivalent to proving that
y+z
2x+z+x
2y+x+y
2z≥3a n d2x
y+z+2y
z+x+2z
x+y≥3.
For the ﬁrst inequality use the fact thaty
x+x
y≥2 and for the second inequality
use Nesbitt’s inequality.
Solution 2.32. Since in triangles with the same base, the ratio between its altitudes
is equal to the ratio of theirs areas, we have that
PQ
AD+PR
BE+PS
CF=(PBC)
(ABC)+(PCA)
(ABC)+(PAB)
(ABC)=(ABC)
(ABC)=1.
Use inequality (2.3) of Section 2.3.
Solution 2.33. (i) Recall that ( S1+S2+S3)(1
S1+1
S2+1
S3)≥9.
(ii) The non-common vertices of the triangles form a hexagon which is divided
into 6 triangles S1,S2,S3,T1,T2,T3,w h e r e SiandTihave one common angle.

146 Solutions to Exercises and Problems
Using the formula for the area that is related to the sine of the angle, prove that
S1S2S3=T1T2T3. After this, use the AM-GM inequality as follows:
S/parenleftbigg1
S1+1
S2+1
S3/parenrightbigg
≥(S1+S2+S3+T1+T2+T3)/parenleftbigg1
S1+1
S2+1
S3/parenrightbigg
≥186√
S1S2S3T1T2T3
3√
S1S2S3=1 8.
The equality holds when the point Ois the centroid of the triangle and the lines
through Oare the medians of the triangle; in this case S1=S2=S3=T1=T2=
T3=1
6S.
Solution 2.34. IfP=Gis the centroid, the equality is evident sinceAG
GL=BG
GM=
CG
GN=2 .
On the other hand, ifAP
PL+BP
PM+CP
PN=6 ,w eh a v eAL
PL+BM
PM+CN
PN=9 .I t
is not diﬃcult to see thatPL
AL=(PBC )
(ABC ),PM
BM=(PCA )
(ABC )andPN
CN=(PAB )
(ABC ), therefore
PL
AL+PM
BM+PN
CN= 1. This implies that
/parenleftbiggAL
PL+BM
PM+CN
PN/parenrightbigg/parenleftbiggPL
AL+PM
BM+PN
CN/parenrightbigg
=9.
By inequality (2.3), the equality above holds only in the case whenAL
PL=BM
PM=
CN
PN= 3, which implies that Pis the centroid.
Solution 2.35. (i) It is known that HD=DD/prime,HE=EE/primeandHF=FF/prime,w h e r e
His the orthocenter.20Thus, the solution follows from part (i) of Example 2.3.4.
(ii) SinceAD/prime
AD=AD +DD/prime
AD=1+HD
AD, we also have, after looking at the solution
to Example 2.3.4, thatAD/prime
AD+BE/prime
BE+CF/prime
CF=1+HD
AD+1+HE
BE+1+HF
CF=4 .
Since/parenleftbigAD
AD/prime+BE
BE/prime+CF
CF/prime/parenrightbig/parenleftBig
AD/prime
AD+BE/prime
BE+CF/prime
CF/parenrightBig
≥9, we have the result.
Solution 2.36. As it has been mentioned in the proof of Example 2.3.5, the length
of the internal bisector of angle Asatisﬁes
l2
a=bc/parenleftBigg
1−/parenleftbigga
b+c/parenrightbigg2/parenrightBigg
=4bc
(b+c)2(s(s−a)).
Since 4 bc≤(b+c)2, it follows that l2
a≤s(s−a)a n d lalb≤s/radicalbig
(s−a)(s−b)≤
s(s−a)+(s−b)
2=sc
2.
Therefore, lalblc≤s/radicalbig
s(s−a)(s−b)(s−c)=s(sr),lalb+lblc+lcla≤
s/parenleftbiga+b+c
2/parenrightbig
=s2andl2
a+l2
b+l2
c≤s(s−a)+s(s−b)+s(s−c)=s2.
20See [6, page 85] or [9, page 37].

4.2 Solutions to the exercises in Chapter 2 147
Solution 2.37. Letα=∠AMB ,β=∠BNA,γ=∠APC,a n dl e t( ABC)b et h e
area of ABC.W eh a v e
(ABC)=1
2a·AMsinα=abc
4R.
Hence,bc
AM=2Rsinα. Similarly,ca
BN=2Rsinβandab
CP=2Rsinγ.T h u s ,
bc
AM+ca
BN+ab
CP=2R(sinα+s i nβ+s i nγ)≤6R.
Equality is attained if M,NandPare the feet of the altitudes.
Solution 2.38. LetA1,B1,C1be the midpoints of the sides BC,CA,AB,r e –
spectively, and let B2,C2be the reﬂections of A1with respect to ABandCA,
respectively. Also, consider Das the intersection of ABwithA1B2andEthe
intersection of CAwithA1C2. Then,
2DE=B2C2≤C2B1+B1C1+C1B2=A1B1+B1C1+C1A1=s.
Use the fact that the quadrilateral A1DAE is inscribed on a circle of diameter
AA 1and the sine law on ADE, to deduce that DE=AA 1sinA=masinA. Then,
s≥2DE=2masinA=2maa
2R=am a
R,t h a ti s , ama≤sR. Similarly, we have
thatbmb≤sRandcmc≤sR.
Solution 2.39. The inequality is equivalent to 8( s−a)(s−b)(s−c)≤abc,w h e r e
sis the semiperimeter.
Since ( ABC)=sr=abc
4R=/radicalbig
s(s−a)(s−b)(s−c), where randRdenote
the inradius and the circumradius of ABC, respectively; we only have to prove
that 8 sr2≤abc,t h a ti s ,8 sr2≤4Rrs, which is equivalent to 2 r≤R.
Solution 2.40. The area of a triangle ABC satisﬁes the equalities ( ABC)=abc
4R=
(a+b+c)r
2, therefore1
ab+1
bc+1
ca=1
2Rr≥1
R2,w h e r e Randrdenote the circumradius
and the inradius, respectively.
Solution 2.41. Use Exercise 2.40 and the sine law.
Solution 2.42. Use that21sinA
2=/radicalBig
(s−b)(s−c)
bc,w h e r e sdenotes the semiperimeter
of the triangle ABC, and similar expressions for sinB
2and sinC
2,t os e et h a t
sinA
2sinB
2sinC
2=(s−a)(s−b)(s−c)
abc=sr2
abc=r
4R≤1
8,
where Randrare the circumradius and the inradius of ABC, respectively.
21Notice that sin2A
2=1−cosA
2=1−b2+c2−a2
2bc
2=a2−(b−c)2
4bc=(s−b)(s−c)
bc.

148 Solutions to Exercises and Problems
Solution 2.43. From inequality (2.3), we know that
(a+b+c)/parenleftbigg1
a+1
b+1
c/parenrightbigg
≥9.
Since a+b+c≤3√
3R,w eh a v e
1
a+1
b+1
c≥√
3
R. (4.4)
Applying, once more, inequality (2.3), we get
1
3/parenleftBigπ
2A+π
2B+π
2C/parenrightBig
≥3
2
π(A+B+C)=3
2. (4.5)
Letf(x)=l o gπ
2x,s i n c e f/prime/prime(x)=1
x2>0,fis convex. Using Jensen’s inequality,
we get
1
3/parenleftBig
logπ
2A+l o gπ
2B+l o gπ
2C/parenrightBig
≥log/bracketleftbigg1
3/parenleftBigπ
2A+π
2B+π
2C/parenrightBig/bracketrightbigg
.
Applying (4.5) and the fact that log xis a strictly increasing function, we obtain
1
3/parenleftBig
logπ
2A+l o gπ
2B+l o gπ
2C/parenrightBig
≥log3
2. (4.6)
We can suppose that a≤b≤c, which implies A≤B≤C. Therefore1
a≥1
b≥1
c
and logπ
2A≥logπ
2B≥logπ
2C. Using Tchebyshev’s inequality,
1
alogπ
2A+1
blogπ
2B+1
clogπ
2C≥/parenleftbigg1
a+1
b+1
c/parenrightbigg/parenleftbigglogπ
2A+l o gπ
2B+l o gπ
2C
3/parenrightbigg
.
Therefore, using (4.4) and (4.6) leads us to
1
alogπ
2A+1
blogπ
2B+1
clogπ
2C≥√
3
Rlog3
2.
Now, raising the expresions to the appropriate powers and taking the reciprocals,
we obtain the desired inequality. In all the above inequalities, the equality holds
if and only if a=b=c(this means, equality is obtained if and only if the triangle
is equilateral).
Solution 2.44. By the sine law, it follows that
sinA
a=sinB
b=sinC
c=1
2R,

4.2 Solutions to the exercises in Chapter 2 149
where a,b,care the lengths of the sides of the triangle and Ris the circumradius
of the triangle. Thus,
sin2A+s i n2B+s i n2C=a2
4R2+b2
4R2+c2
4R2
=1
4R2(a2+b2+c2)
≤1
4R2·9R2=9
4,
where the inequality follows from Leibniz’s inequality.
Solution 2.45. Use Leibniz’s inequality and the fact that the area of a triangle is
given by ( ABC)=abc
4R.
Solution 2.46. We note that the incircle of ABC is the circumcircle of DEF.
Applying Leibniz’s inequality to DEF,w eg e t
EF2+FD2+DE2≤9r2,
where ris the inradius of ABC. On the other hand, using Theorem 2.4.3 we obtain
s2≥27r2, hence
EF2+FD2+DE2≤s2
3.
Solution 2.47.
a2
hbhc+b2
hcha+c2
hahb=a2bc+b2ca+c2ab
4(ABC)2=abc(a+b+c)
4(ABC)2
=abc(a+b+c)
4abc
4R(a+b+c)r
2=2R
r≥4.
Solution 2.48. Remember that sin2A
2=1−cosA
2and use that cos A+c o s B+
cosC≤3
2(see Example 2.5.2).
Solution 2.49. Observe that
4√
3(ABC)≤9abc
a+b+c⇔4√
3rs≤9·4Rrs
2s⇔2√
3s≤9R⇔2s
3√
3≤R.
The last inequality was proved in Theorem 2.4.3.
Solution 2.50. Use the previous exercise and the inequality between the harmonic
mean and the geometric mean,
3
1
ab+1
bc+1
ca≤3√
a2b2c2.

150 Solutions to Exercises and Problems
Solution 2.51. Use the previous exercise and the AM-GM inequality,
3√
a2b2c2≤a2+b2+c2
3.
Solution 2.52. First, observe that if s=a+b+c
2,t h e n
a2+b2+c2−(a−b)2−(b−c)2−(c−a)2=
=a2−(b−c)2+b2−(c−a)2+c2−(a−b)2
=4{(s−b)(s−c)+(s−c)(s−a)+(s−a)(s−b)}.
Hence, if x=s−a,y=s−b,z=s−c, then the inequality is equivalent to
√
3/radicalbig
xyz(x+y+z)≤xy+yz+zx.
Squaring and simplifying the last inequality, we get
xyz(x+y+z)≤x2y2+y2z2+z2x2.
This inequality can be deduced using Cauchy-Schwarz’s inequality with
(xy,yz,zx )a n d( zx,xy,yz ).
Solution 2.53. Use Exercise 2.50 and the inequality 33/radicalbig
(ab)(bc)(ca)≤ab+bc+ca.
Solution 2.54. Note that
3(a+b+c)abc
ab+bc+ca≥9abc
a+b+c⇔(a+b+c)2≥3(ab+bc+ca)
⇔a2+b2+c2≥ab+bc+ca,
now, use Exercise 2.49.
Solution 2.55. Using (2.5), (2.6) and (2.7) we can observe that a2+b2+c2+4abc=
1
2−2r2.
Solution 2.56. Observe the relationships used in the proof of Exercise 2.39,
(b+c−a)(c+a−b)(a+b−c)
abc=8(s−a)(s−b)(s−c)
abc
=8s(s−a)(s−b)(s−c)
4Rs(abc
4R)
=8(rs)2
4Rs(rs)=2r
R.

4.2 Solutions to the exercises in Chapter 2 151
Solution 2.57. Observe that
a2
b+c−a+b2
c+a−b+c2
a+b−c=1
2/parenleftbigga2
s−a+b2
s−b+c2
s−c/parenrightbigg
=1
2/parenleftbiggsa
s−a−a+sb
s−b−b+sc
s−c−c/parenrightbigg
=s
2/parenleftbigga
s−a+b
s−b+c
s−c/parenrightbigg
−s
=s
2/bracketleftbigg(a+b+c)s2−2(ab+bc+ca)s+3abc
(s−a)(s−b)(s−c)/bracketrightbigg
−s
=s
2/bracketleftbigg2s3−2s(s2+r2+4rR)+3 ( 4 Rrs)
r2s/bracketrightbigg
−s
=2s(R−r)
r≥2s(R−R
2)
r≥3√
3rR
r=3√
3R,
the last two inequalities follow from the fact that R≥2r(which implies that
−r≥−R
2)a n df r o m s≥3√
3r, respectively.
Solution 2.58. Start on the side of the equations which expresses the relationship
between the τ’s and perform the operations.
Solution 2.59. Ifx1,1−x1,x2,1−x2,…are the lengths into which each side
is divided for the corresponding point, we can deduce that a2+b2+c2+d2=/summationtext(x2
i+( 1−xi)2). Prove that1
2≤2(xi−1
2)2+1
2=x2i+( 1−xi)2≤1.
For part (ii), the inequality on the right-hand side follows from the triangle
inequality. For the one on the left-hand side, use reﬂections on the sides, as you
can see in the ﬁgure.
ba
cd
Solution 2.60. This is similar to part (ii) of the previous problem.

152 Solutions to Exercises and Problems
Solution 2.61. IfABC is the triangle and DEFGHI is the hexagon with DE,
FG,HIparallel to BC,AB,CA, respectively, we have that the perimeter of the
hexagon is 2( DE+FG+HI). Let X,Y,Zbe the tangency points of the incircle
with the sides BC,CA,AB, respectively, and let p=a+b+cbe the perimeter of
the triangle ABC.S e tx=AZ=AY,y=BZ=BXandz=CX=CY,t h e n
we have the relations
DE
a=AE+ED+DA
p=2x
p.
Similarly, we have the other relations
FG
c=2z
p,HI
b=2y
p.
Therefore,
p(DEFGHI )=4(xa+yb+zc)
p=4(a(s−a)+b(s−b)+c(s−c))
2s
=4((a+b+c)s−(a2+b2+c2))
2s
=2 (a+b+c)−4(a2+b2+c2)
(a+b+c),
buta2+b2+c2≥1
3(a+b+c)(a+b+c) by Tchebyshev’s inequality. Thus,
p(DEFGHI )≤2(a+b+c)−4
3(a+b+c)=2
3(a+b+c).
Solution 2.62. Take the circumcircle of the equilateral triangle with side length 2.
The circles with centers the midpoints of the sides of the triangle and radii 1 cover
a circle of radius 2. If a circle of radius greater than2√
3
3is covered by three circles
of radius 1, then one of the three circles covers a chord of length greater than 2.
Solution 2.63. Take the acute triangle with sides of lengths 2 r1,2r2and 2r3,i fi t
exists. Its circumradius is the solution. If the triangle does not exist, the maximum
radius between r1,r2andr3is the answer.
Solution 2.64. We need two lemmas.
Lemma 1. If a square of side-length alies inside a rectangle of sides candd,t h e n
a≤min{c, d}.
cd
a

4.2 Solutions to the exercises in Chapter 2 153
Through the vertices of the square draw parallel lines to the sides of the
rectangle in such a way that those lines enclose the square as in the ﬁgure. Since
the parallel lines form a square inside the rectangle and such a square containsthe original square, we have the result.
Lemma 2. The diagonal of a square inscribed in a right triangle is less than or
equal to the length of the internal bisector of the right angle.
LetABC be the right triangle with hypotenuse CAand let PQRS be the
inscribed square.
It can be assumed that the vertices PandQbelong to the legs of the right
triangle (otherwise, translate the square) and let Obe the intersection point of
the diagonals PRandQS.
A
B CRS
T
P
QO
O/prime
V
Since BQOP is cyclic (∠ B=∠O=9 0◦), it follows that ∠QBO =∠QPO =
45◦,t h e n Obelongs to the internal bisector of ∠B.L e tTbe the intersection of
BOwithRS,t h e n ∠QBT =∠QST =4 5◦, therefore BQTS is cyclic and the
center O/primeof the circumcircle of BQTS is the intersection of the perpendicular
bisectors of SQandBT. But the perpendicular bisector of SQisPR, hence the
point O/primebelongs to PR,a n di f Vis the midpoint of BT,w eh a v et h a t VO O/primeis a
right triangle. Since O/primeO>O/primeV, then the chords SQandBTsatisfy SQ < BT ,
and the lemma follows.
Let us ﬁnish now the proof of the problem. Let ABCD be the square of side
1a n dl e t lbe a line that separates the two squares. If lis parallel to one of the
sides of the square ABCD , then Lemma 1 applies. Otherwise, lintersects every
line that determines a side of the square ABCD . Suppose that Ais the farthest
vertex from l.

154 Solutions to Exercises and Problems
a
bA
B
CD
G HEF
Iflintersects the sides of ABCD inE,F,G,Has in the ﬁgure, we have, by
Lemma 2, that the sum of the lengths of the diagonals of the small squares is less
than or equal to AC,t h a ti s ,√
2(a+b)≤√
2, then the result follows.
Solution 2.65. Ifα,β,γare the central angles which open the chords of length a,
b,c, respectively, we have that a=2s i nα
2,b=2s i nβ
2andc=2s i nγ
2. Therefore,
abc=8s i nα
2sinβ
2sinγ
2≤8s i n3/parenleftbiggα+β+γ
6/parenrightbigg
=8s i n3(30◦)=1,
where the inequality follows from Exercise 1.81.
Solution 2.66. The ﬁrst observation that we should make is to check that the
diagonals are parallel to the sides. Let Xbe the point of intersection between the
diagonals ADandCE. Now, the pentagon can be divided into
(ABCDE )=(ABC)+(ACX)+(CDE)+(EAX).
Since ABCX is a parallelogram, we have ( ABC)=(CXA)=(CDE). Let a=
(CDX )=(EAX)a n d b=(DEX ), then we geta
b=AX
XD=(CXA )
(CDX )=a+b
a,t h a t
is,a
b=1+√
5
2. Now we have all the elements to ﬁnd ( ABCDE ).
Solution 2.67. Prove that sr=s1R=(ABC), where s1is the semiperimeter of
the triangle DEF. To deduce this equality, it is suﬃcient to observe that the radii
OA,OBandOCare perpendicular to EF,FDandDE, respectively. Use also
thatR≥2r.
Solution 2.68. Suppose that the maximum angle is Aa n dt h a ti ts a t i s ﬁ e s6 0◦≤
A≤90◦, then the lengths of the altitudes hbandhcare also less than 1. Now, use
t h ef a c tt h a t( ABC)=hbhc
2s i nAand that√
3
2≤sinA≤1. The obtuse triangle case
is easier.

4.2 Solutions to the exercises in Chapter 2 155
Solution 2.69. LetABCD be the quadrilateral with sides of length a=AB,
b=BC,c=CDandd=DA.
(i) (ABCD )=(ABC)+(CDA)=absinB
2+cdsinD
2≤ab+cd
2.
(ii) If ABCD is the quadrilateral mentioned with sides of length a,b,candd,
consider the triangle BC/primeDwhich results from the reﬂection of DCB with respect
to the perpendicular bisector of side BD. The quadrilaterals ABCD andABC/primeD
have the same area but the second one has sides of length a,c,bandd,i nt h i s
order. Now use (i).
(iii) (ABC)≤ab
2,(BCD)≤bc
2,(CDA)≤cd
2and (DAB)≤da
2.
Solution 2.70. In Example 2.7.6 we proved that
PA·PB·PC≥R
2r(pa+pb)(pb+pc)(pc+pa).
Use the AM-GM inequality.
Solution 2.71. (i)PA2
pbpc+PB2
pcpa+PC2
papb≥33/radicalBig
PA2
pbpcPB2
pcpaPC2
papb≥33/radicalBig
/parenleftbig4R
r/parenrightbig2≥12.
(ii)PA
pb+pc+PB
pc+pa+PC
pa+pb≥33/radicalBig
PA
pb+pcPB
pc+paPC
pa+pb≥33/radicalBig
R
2r≥3.
(iii)PA
√
pbpc+PB
√
pcpa+PC
√
papb≥33/radicalBig
PA
√
pbpcPB
√
pcpaPC
√
papb≥33/radicalBig
4R
r≥6.
For the last inequalities in (i) and (iii), we have used Exercise 2.70. For the
last inequality in (ii), we have resorted to Example 2.7.6.
(iv) Proceed as in Example 2.7.5, that is, apply inversion in a circle with center P
and radius d(arbitrary, for instance d=pb). Let A/prime,B/prime,C/primebe the inverses of A,
B,C, respectively. Let p/prime
a,p/prime
b,p/prime
cbe the distances from Pto the sides B/primeC/prime,C/primeA/prime,
A/primeB/prime, respectively.
Let us prove that p/prime
a=paPB/prime·PC/prime
d2.W eh a v e
p/prime
aB/primeC/prime=2 (PB/primeC/prime)=PB/prime·PC/prime·B/primeC/prime
PA/prime
1=paPB/prime·PC/prime·B/primeC/prime
d2,
where A/prime1is the inverse of A1, the projection of PonBC. Similarly, p/prime
b=pbPC/prime·PA/prime
d2
andp/prime
c=pcPA/prime·PB/prime
d2.
The Erd˝ os-Mordell inequality, applied to the triangle A/primeB/primeC/prime, guarantees us
thatPA/prime+PB/prime+PC/prime≥2(p/primea+p/prime
b+p/prime
c).
Now, since PA·PA/prime=PB·PB/prime=PC·PC/prime=d2, after substitution we get
1
PA+1
PB+1
PC≥2/parenleftBigpa
PB·PC+pb
PC·PA+pc
PC·PA/parenrightBig
and this inequality is equivalent to
PB·PC+PC·PA+PA·PB≥2(paPA+pbPB+pcPC).
Finally, to conclude use example 2.7.4.

156 Solutions to Exercises and Problems
Solution 2.72. IfPis an interior point or a point on the perimeter of the triangle
ABC, see the proof of Theorem 2.7.2.
Ifhais the length of the altitude from vertex A, we have that the area of the
triangle ABC satisﬁes 2( ABC)=aha=apa+bpb+cpc.
Since ha≤PA+pa(even if pa≤0, that is, if Pis a point on the outside of
the triangle, on a diﬀerent side of BCthanA), and because the equality holds if P
is exactly on the segment of the altitude from the vertex A, therefore aPA+apa≥
aha=apa+bpb+cpc, hence aPA≥bpb+cpc.
This inequality can be applied to triangle AB/primeC/primesymmetric to ABC with
respect to the internal angle bisector of A,w h e r e aPA≥cpb+bpc, with equality
when APpasses through the point O.
Similarly, bPB≥apc+cpaandcPC≥apb+bpa, therefore
PA+PB+PC≥/parenleftbiggb
c+c
b/parenrightbigg
pa+/parenleftBigc
a+a
c/parenrightBig
pb+/parenleftbigga
b+b
a/parenrightbigg
pc.
We have the equality when Pis the circumcenter O.
Second solution. LetL,MandNbe the feet of the perpendicular from point
Pto the sides BC,CAandAB, respectively. Let HandGbe the orthogonal
projections of BandC, respectively, over the segment MN.T h e n BC≥HG=
HN+NM+MG.
Since∠BNH =∠ANM =∠APM , the right triangles BNH andAPM are
similar, therefore HN=PM
PABN.In an analogous way we get MG=PN
PACM.
Applying Ptolemy’s theorem to AMPN ,w eo b t a i n PA·MN=AN·PM+
AM·PN, hence
MN=AN·PM+AM·PN
PA,
f r o mt h e r ew eg e t
BC≥PM
PABN+AN·PM+AM·PN
PA+PN
PACM.
Therefore,
BC·PA≥PM·AB+PN·CA.
Then, PA≥pbc
a+pcb
a. Similarly for the other two inequalities.
Solution 2.73. Take a sequence of reﬂections of the quadrilateral ABCD ,a ss h o w n
in the ﬁgure.

4.2 Solutions to the exercises in Chapter 2 157
ABCD
A/primeA/prime/prime
B/primeB/prime/prime
D/prime
C/prime
S
PQRR/primeP/prime/prime
S/prime/prime
R/prime/prime
Note that the perimeter of PQRS is the sum of the lengths of the piecewise line
PQR/primeS/prime/primeP/prime/prime.N o t ea l s ot h a t A/prime/primeB/prime/primeis parallel to ABand that the shortest distance
isAA/prime/primeas can be seen if we project Oon the sides of the quadrilateral.
Solution 2.74. F i r s tn o t et h a t( DEF)=(ABC)−(AFE)−(FBD)−(EDC).
Ifx=BD,y=CE,z=AF,a−x=DC,b−y=EAandc−z=FB,w e
have
(AFE)
(ABC)=z(b−y)
cb,(FBD)
(ABC)=x(c−z)
acand(EDC)
(ABC)=y(a−x)
ba.
Therefore,
(DEF)
(ABC)=1−z
c/parenleftBig
1−y
b/parenrightBig
−x
a/parenleftBig
1−z
c/parenrightBig
−y
b/parenleftBig
1−x
a/parenrightBig
=/parenleftBig
1−x
a/parenrightBig/parenleftBig
1−y
b/parenrightBig/parenleftBig
1−z
c/parenrightBig
+x
a·y
b·z
c=2x
a·y
b·z
c.
The last equality follows from the fact thatx
a−x·y
b−y·z
c−z= 1 which is guaranteed
because the cevians occur. Now, the last product is maximum whenx
a=y
b=
z
c, and since the segments concur the common value is1
2.T h u s Pmust be the
centroid.
Solution 2.75. Ifx=PD,y=PEandz=PF, we can deduce that 2( ABC)=
ax+by+cz. Using the Cauchy-Schwarz inequality,
(a+b+c)2≤/parenleftbigga
x+b
y+c
z/parenrightbigg
(ax+by+cz).
Thena
x+b
y+c
z≥(a+b+c)2
2(ABC )and the equality holds when x=y=z,t h a ti s ,w h e n
Pis the incenter.

158 Solutions to Exercises and Problems
Solution 2.76. First, observe that BD2+CE2+AF2=DC2+EA2+FB2,w h e r e
BD2−DC2=PB2−PC2and similar relations have been used.
Now, ( BD+DC)2=a2, hence BD2+DC2=a2−2BD·DC. Similarly for
the other two sides. Thus, BD2+DC2+CE2+AE2+AF2+FB2=a2+b2+
c2−2(BD·DC+CE·AE+AF·FB).
In this way, the sum is minimum when ( BD·DC+CE·AE+AF·FB)i s
maximum. But BD·DC≤/parenleftbigBD +DC
2/parenrightbig2=/parenleftbiga
2/parenrightbig2and the maximum is attained when
BD=DC. Similarly, CE=EAandAF=FB, therefore Pis the circumcenter.
Solution 2.77. Since3/radicalbig
(aPD)(bPE )(cPF )≤aPD +bPE +cPF
3=2(ABC )
3,w ec a n
deduce that PD·PE·PF≤8
27(ABC )3
abc. Moreover, the equality holds if and only
ifaPD=bPE=cPF.
Butc·PF=b·PE⇔(ABP)=( CAP)⇔Pis on the median AA/prime.
Similarly, we can see that Pis on the other medians, thus Pis the centroid.
Solution 2.78. Using the technique for proving Leibniz’s theorem, verify that
3PG2=PA2+PB2+PC2−1
3(a2+b2+c2), where Gis the centroid of ABC.
Therefore, the optimal point must be P=G.
Solution 2.79. The quadrilateral APMN is cyclic and it is inscribed in the circle of
diameter AP.T h ec h o r d MNalways opens the angle A(or 180◦−∠A), therefore
the length of MNwill depend proportionally on the radius of the circumscribed
circle to APMN . The biggest circle will be attained when the diameter APis the
biggest possible. This happens when Pis diametrally opposed to A.I nt h i sc a s e M
NA
P
C BM
andNcoincide with BandC, respectively. Therefore the maximum chord MN
isBC.
Solution 2.80. The circumcircle of DEF is the nine-point circle of ABC, therefore
it intersects also the midpoints of the sides of ABC and goes through L,M,N,
the midpoints of AH,BH,CH, respectively.

4.2 Solutions to the exercises in Chapter 2 159
COH
MF
A/primeDA
N
B∠AEL
Note that t2
a=AL·AD,t h e n
/summationdisplayt2a
ha=/summationdisplayAL·AD
AD=/summationdisplay
AL=/summationdisplay
OA/prime
=/summationdisplay
RcosA≤3RcosA+B+C
3=3Rcos60◦=3
2R.
Observe that we can prove a stronger result/summationtextt2
a
ha=R+r, using the fact that
cosA+c o s B+c o s C=r
R+ 1. See Lemma 2.5.2.
Solution 2.81. (i) Notice that
pa
ha+pb
hb+pc
hc=apa
aha+bpb
bhb+cpc
chc
=2(PBC)+2 ( PCA)+2 ( PAB)
2(ABC)=1.
Now use the fact that/parenleftbiggpa
ha+pb
hb+pc
hc/parenrightbigg/parenleftbiggha
pa+hb
pb+hc
pc/parenrightbigg
≥9.
(ii) Using the AM-GM inequality, we have
27/parenleftbiggpa
hapb
hbpc
hc/parenrightbigg
≤/parenleftbiggpa
ha+pb
hb+pc
hc/parenrightbigg3
=1 ,
where the last equality follows from (i).
(iii) Let x=(PBC),y=(PCA)a n d z=(PAB). Observe that a(ha−pa)=
aha−apa=2 (y+z)≥4√
yz. Similarly, we have that b(hb−pb)≥4√
zxy
c(hc−pc)≥4√
xy. Then,
a(ha−pa)b(hb−pb)c(hc−pc)≥64xyz =8 (apabpbcpc).
Therefore, ( ha−pa)(hb−pb)(hc−pc)≥8papbpc.

160 Solutions to Exercises and Problems
Solution 2.82. Assume that a<b<c , then of all the altitudes of ABC,ADis the
longest. If Eis the projection of IonAD, it is enough to prove that AE≥AO=R.
Remember that the internal bisector of ∠Ais also the internal bisector of ∠EAO.
IfIis projected on E/primein the diameter AA/prime,t h e n AE=AE/prime. Now prove that
AE/prime≥AO,b yp r o v i n gt h a t Iis inside the acute triangle COF,w h e r e Fis the
intersection of AA/primewithBC.
To see that COF is an acute triangle, use that the angles of ABC satisfy
∠A<∠B<∠C,s ot h a t1
2∠B<90◦−∠A,1
2∠C<90◦−∠A. Use also that
∠COF =∠A+∠C−∠B<90◦.
Solution 2.83. LetABC be a triangle with sides of lengths a,bandc.U s i n g
Heron’s formula to calculate the area of the triangle, we have that
(ABC)=/radicalbig
s(s−a)(s−b)(s−c),where s=a+b+c
2. (4.7)
Ifsandcare ﬁxed, then s−cis also ﬁxed. Then the product 16( ABC)2is
maximum when ( s−a)(s−b) is maximum, that is, if s−a=s−b,w h i c hi s
equivalent to a=b. Therefore the triangle is isosceles.
Solution 2.84. LetABC be a triangle with sides of length a,bandc.S i n c et h e
perimeter is ﬁxed, the semi-perimeter is also ﬁxed. Using (4.7), we have that
16(ABC)2is maximum when (s −a)(s−b)(s−c) is maximum. The product of
these three numbers is maximum when ( s−a)=(s−b)=(s−c), that is, when
a=b=c. Therefore, the triangle is equilateral.
Solution 2.85. Ifa,b,care the lengths of the sides of the triangle, observe that
a+b+c=2R(sin∠A+sin∠B+sin∠C)≤6Rsin/parenleftbig∠A+∠B+∠C
3/parenrightbig
, since the function
sinxis concave. Moreover, equality holds when sin ∠A=s i n∠B=s i n∠C.
Solution 2.86. The inequality ( lm+mn+nl)(l+m+n)≥a2l+b2m+c2nis
equivalent to
l2+m2−c2
lm+m2+n2−b2
mn+n2+l2−a2
nl+3≥0
⇔cos∠APB +c o s∠BPC +c o s∠CPA +3
2≥0.
N o wu s et h ef a c tt h a tc o s α+c o s β+c o s γ+3
2≥0i se q u i v a l e n tt o( 2 c o sα+β
2+
cosα−β
2)2+s i n2(α−β
2)≥0.
Solution 2.87. Consider the Fermat point Fand let p1=FA,p2=FBand
p3=FC, then observe ﬁrst that ( ABC)=1
2(p1p2+p2p3+p3p1)sin120◦=√
3
4(p1p2+p2p3+p3p1). Also,
a2+b2+c2=2p2
1+2p22+2p23−2p1p2cos120◦−2p2p3cos120◦−2p3p1cos120◦
=2 (p2
1+p22+p23)+p1p2+p2p3+p3p1.

4.2 Solutions to the exercises in Chapter 2 161
Now, using the fact that x2+y2≥2xy, we can deduce that a2+b2+c2≥
3(p1p2+p2p3+p3p1)=3/parenleftbig4
3√
3(ABC)/parenrightbig
. Then, a2+b2+c2≥4√
3(ABC).
Moreover, the equality a2+b2+c2=4√
3(ABC)h o l d sw h e n p2
1+p22+p23=
p1p2+p2p3+p3p2,t h a ti s ,w h e n p1=p2=p3or, equivalently, when the triangle
is equilateral.
Solution 2.88. Leta,b,cbe the lengths of the sides of the triangle ABC.I nt h e
same manner as we proceeded in the previous exercise, deﬁne p1=FA,p2=FB
andp3=FC. From the solution of the previous exercise we know that
4√
3(ABC)=3 ( p1p2+p2p3+p3p1).
Thus, we only need to prove that
3(p1p2+p2p3+p3p1)≤(p1+p2+p3)2,
but this is equivalent to p1p2+p2p3+p3p1≤p2
1+p22+p23, which is Exercise 1.27.
Solution 2.89. As in the Fermat problem there are two cases, when in ABC all
angles are less than 120◦or when there is an angle greater than 120◦.
In the ﬁrst case the minimum of PA+PB+PCisCC/prime,w h e r e C/primeis the
image of Awhen we rotate the ﬁgure in a positive direction through an angle of
60◦having Bas the center. Using the cosine law, we obtain
(CC/prime)2=b2+c2−2bccos(A+6 0◦)
=b2+c2−bccosA+bc√
3sinA
=1
2(a2+b2+c2)+2√
3(ABC).
Now, use the fact that a2+b2+c2≥4√
3(ABC)t oo b t a i n( CC/prime)2≥4√
3(ABC).
Applying Theorem 2.4.3 we have that ( ABC)≥3√
3r2, therefore ( CC/prime)2≥36r2.
When ∠A≥120◦, the point that solves Fermat-Steiner problem is the point
A,t h e n PA+PB+PC≥AB+AC=b+c. It suﬃces to prove that b+c≥6r.
Moreover, we can use the fact that b=x+z,c=x+yandr=/radicalBig
xyz
x+y+z.
Second solution. It is clear that PA+pa≥ha,w h e r e pais the distance from
PtoBCandhais the length of the altitude from A.T h e n ha+hb+hc≤
(PA+PB+PC)+(pa+pb+pc)≤3
2(PA+PB+PC), where the last inequality
follows from Erd˝ os-Mordell’s theorem.
Now using Exercise 1.36 we have that 9 ≤(ha+hb+hc)(1
ha+1
hb+1
hc)=
(ha+hb+hc)(1
r). Therefore, 9 r≤ha+hb+hc≤3
2(PA+PB+PC)a n dt h e
result follows.
Solution 2.90. First, we note that (A 1B1C1)=1
2A1B1·A1C1·sin∠B1A1C1.Since
PB 1CA 1is a cyclic quadrilateral with diameter PC, applying the sine law leads
us to A1B1=PCsinC. Similarly, A1C1=PBsinB.

162 Solutions to Exercises and Problems
CallQthe intersection of BPwith the circumcircle of triangle ABC,t h e n
∠B1A1C1=∠QCP. In fact, since PB 1CA 1is a cyclic quadrilateral we have
∠B1CP=∠B1A1P. Similarly, ∠C1BP=∠C1A1P.T h e n ∠B1A1C1=∠B1A1P
+∠C1A1P=∠B1CP+∠C1BP, but∠C1BP=∠ABQ =∠ACQ. Therefore,
∠B1A1C1=∠B1CP+∠ACQ =∠QCP.
Once again, the sine law guarantees thatsin∠QCP
sin∠BQC=PQ
PC.
(A1B1C1)=1
2A1B1·A1C1sin∠B1A1C1
=1
2PB·PCsinBsinCsin∠QCP
=1
2PB·PC·sinBsinCPQ
PCsin∠BQC
=1
2PB·PQ·sinAsinBsinC
=(R2−OP2)(ABC )
4R2.
The last equality holds true because the power of the point Pwith respect to
the circumcircle of ABC isPB·PQ=R2−OP2, and because ( ABC)=
2R2sinAsinBsinC. The area of A1B1C1is maximum when P=O,t h a ti s ,
when A1B1C1is the medial triangle.
/Bullet/Bullet
BCA
B1
A1PC1OQ
4.3 Solutions to the problems in Chapter 3
Solution 3.1. Leta=A1A2,b=A1A3andc=A1A4. Using Ptolemy’s theorem
in the quadrilateral A1A3A4A5, we can deduce that ab+ac=bcor, equivalently,
a
b+a
c=1 .

4.3 Solutions to the problems in Chapter 3 163
Since the triangles A1A2A3andB1B2B3are similar,B1B2
B1B3=A1A2
A1A3=a
b
and from there we obtain B1B2=a2
b. Similarly C1C2=a2
c. ThereforeSB+SC
SA=
a2
b2+a2
c2=a2c2+a2b2
b2c2=b2+c2
(b+c)2>(b+c)2
2(b+c)2=1
2. The third equality follows from
ab+ac=bcand the inequality follows from inequality (1.11). The inequality is
strict since b/negationslash=c.
Note thata2
b2+a2
c2=/parenleftbiga
b+a
c/parenrightbig2−2a2
bc=1−2a2
bc.
The sine law applied to the triangle A1A3A4leads us to
a2
bc=sin2π
7
sin2π
7sin4π
7=sin2π
7
2s i n2π
7sin2π
7cos2π
7
=sin2π
7
2(1−cos22π
7)cos2π
7=sin2π
7
2c o s2π
7(1 + cos2π
7)(1−cos2π
7)
=sin2π
7
4c o s2π
7(1 + cos2π
7)sin2π
7=1
4c o s2π
7(1 + cos2π
7)
>1
4c o sπ
4(1 + cosπ
4)=1
4√
2
2(1 +√
2
2)=√
2−1
2.
Thusa2
b2+a2
c2=1−2a2
bc<1−(√
2−1) = 2 −√
2.
Solution 3.2. Cut the tetrahedron along the edges AD,BD,CDand place it on
the plane of the triangle ABC. The faces ABD,BCD andCAD will have as their
image the triangles ABD 1,BCD 2andCAD 3.O b s e r v et h a t D3,AandD1are
collinear, as are D1,BandD2.M o r e o v e r , Ais the midpoint of D1D3(since both
D1AandD3Aare equal in length to DA), and similarly Bis the midpoint of D1D2.
Then AB=1
2D2D3and by the triangle inequality, D2D3≤CD 3+CD 2=2CD.
Hence AB≤CD, as desired.
Solution 3.3. Letting Sbe the area of the triangle, we have the formulae sin α=2S
bc,
sinβ=2S
ca,s i nγ=2S
abandr=S
s=2S
a+b+c. Using these formulae we ﬁnd that the
inequality to be proved is equivalent to
/parenleftbigga
bc+b
ca+c
ab/parenrightbigg
(a+b+c)≥9,
which can be proved by applying the AM-GM inequality to each factor on the left
side.
Solution 3.4. Suppose that the circles have radii 1. Let Pbe the common point of
the circles and let A,B,Cbe the second intersection points of each pair of circles.
We have to minimize the common area between any pair of circles, which will be
minimum if the point Pis in the interior of the triangle ABC (otherwise, rotate
one circle by 180◦around P, and this will reduce the common area).
The area of the common parts is equal to π−(sinα+s i nβ+s i nγ), where
α,β,γare the central angles of the common arcs of the circles. It is clear that

164 Solutions to Exercises and Problems
γBβαA
γ
Pα
Cβ
α+β+γ= 180◦. Since the function sin xis concave, the minimum is reached when
α=β=γ=π
3, which implies that the centers of the circles form an equilateral
triangle.
Solution 3.5. LetIbe the incenter of ABC, and draw the line through Iperpendic-
ular to IC.L e tD/prime,E/primebe the intersections of this line with BC,CA, respectively.
First prove that ( CDE)≥(CD/primeE/prime) by showing that the area of D/primeDIis greater
than or equal to the area of EE/primeI; to see this, observe that one of the triangles
DD/primeI,EE/primeIlies in the opposite side to Cwith respect to the line D/primeE/prime,i ff o r
instance, it is DD/primeI, then this triangle will have a greater area than the area of
EE/primeI, then the claim.
Now, prove that the area (CD/primeE/prime)i s2r2
sinC;t os e et h i s ,n o t et h a t CI=r
sinC
2
and that D/primeI=r
cosC
2,t h e n
(CD/primeE/prime)=1
2D/primeE/prime·CI=D/primeI·CI=2r2
2s i nC
2cosC
2=2r2
sinC≥2r2.
Solution 3.6. The key is to note that 2 AX≥√
3(AB+BX), which can be deduced
by applying Ptolemy’s inequality (Exercise 2.11) to the cyclic quadrilateral ABXO
that is formed when we glue the triangle ABX to the equilateral triangle AXO of
sideAX, and then observing that the diameter of the circumcircle of the equilateral
triangle is2
√
3AX,t h a ti s , AX(AB+BX)=AX·BO≤AX·2
√
3AX. Hence
2AD=2 (AX+XD)≥√
3(AB+BX)+2XD
≥√
3(AB+BC+CX)+√
3XD
≥√
3(AB+BC+CD).
Solution 3.7. Take the triangle A/primeB/primeC/primeof maximum area between all triangles
that can be formed with three points of the given set of points; then its area
satisﬁes ( A/primeB/primeC/prime)≤1. Construct another triangle ABC that has A/primeB/primeC/primeas its

4.3 Solutions to the problems in Chapter 3 165
medial triangle; this has an area ( ABC)=4 ( A/primeB/primeC/prime)≤4. InABC we can ﬁnd
all the points. Indeed, if some point Qis outside of ABC, it will be in one of the
half-planes determined by the sides of the triangle and opposite to the half-plane
where the third vertex lies. For instance, if Qis in the half-plane determined by
BC, opposite to where Alies, the triangle QB/primeC/primehas greater area than A/primeB/primeC/prime,
a contradiction.
Solution 3.8. LetM=1 +1
2+···+1
n. Let us prove that Mis the desired minimum
value, which is achieved by setting x1=x2=···=xn= 1. Using the AM-GM
inequality, we get xk
k+(k−1)≥kxkfor all k. Therefore
x1+x2
2
2+x33
3+···+xnn
n≥x1+x2−1
2+···+xn−n−1
n=x1+x2+···+xn−n+M.
On the other hand, the arithmetic-harmonic inequality leads us to
x1+x2+···+xn
n≥n
1
x1+1
x1+···+1
xn=1.
We conclude that the given expression is at least n−n+M=M.S i n c e Mis
achieved, it is the desired minimum.
Second solution. Apply the weighted AM-GM inequality to the numbers {xj
j}with
weights/braceleftBig
tj=1
j
/summationtext1
j/bracerightBig
,t og e t
/summationdisplayxjj
j≥/parenleftbigg/summationdisplay1
j/parenrightbigg
(x1x2···xn)1
/summationtext1
j≥/summationdisplay1
j.
The last inequality follows from nn/radicalBig
1
x1···1
xn≤/summationtext1
xj=n.
Solution 3.9. Note that AFE andBDC are equilateral triangles. Let C/primeandF/prime
be points outside the hexagon and such that ABC/primeandDEF/primeare also equilateral
triangles. Since BEis the perpendicular bisector of AD, it follows that C/primeandF/prime
are the reﬂections of CandFon the line BE. Now use the fact that AC/primeBGand
EF/primeDHare cyclic in order to conclude that AG+GB=GC/primeandDH+HE=
HF/prime.
Solution 3.10. Leibniz’s theorem implies OG2=R2−1
9(a2+b2+c2). Since
rs=abc
4R, we can deduce that 2 rR=abc
a+b+c.T h e nw eh a v et op r o v et h a t abc≤
(a+b+c)
3(a2+b2+c2)
3, for which we can use the AM-GM inequality.
Solution 3.11. The left-hand side of the inequality follows from
/radicalbig
1+x0+x1+···+xi−1√
xi+···+xn≤1
2(1 +x0+···+xn)=1.

166 Solutions to Exercises and Problems
For the right-hand side consider θi=a r c s i n( x0+···+xi)f o ri=0,…,n .N o t e
that
/radicalbig
1+x0+···+xi−1√
xi+···+xn=/radicalbig
1+s i n θi−1/radicalbig
1−sinθi−1
=c o s θi−1.
It is left to prove that/summationtextsinθi−sinθi−1
cosθi−1<π
2.B u t
sinθi−sinθi−1=2c o sθi+θi−1
2sinθi−θi−1
2<(cosθi−1)(θi−θi−1).
To show the inequality, use the facts that cos θis a decreasing function and that
sinθ≤θfor 0≤θ≤π
2.T h e n
/summationdisplaysinθi−sinθi−1
cosθi−1</summationdisplay
θi−θi−1=θn−θ0=π
2.
Solution 3.12. If/summationtextn
i=1xi=1 ,t h e n1=(/summationtextni=1xi)2=/summationtextni=1×2
i+2/summationtext
i<jxixj.
Therefore the inequality that we need to prove is equivalent to
1
n−1≤n/summationdisplay
i=1x2i
1−ai.
Use the Cauchy-Schwarz inequality to prove that
/parenleftBiggn/summationdisplay
i=1xi/parenrightBigg2
≤n/summationdisplay
i=1x2i
1−ain/summationdisplay
i=1(1−ai).
Solution 3.13. First prove that/summationtextn
i=1xn+1(xn+1−xi)=(n−1)x2
n+1. The inequality
that we need to prove is reduced to
n/summationdisplay
i=1/radicalbig
xi(xn+1−xi)≤√
n−1xn+1.
Now use the Cauchy-Schwarz inequality with the following two n-sets of real num-
bers: (√
x1,…,√
xn)a n d(√
xn+1−x1,…,√
xn+1−xn).
Solution 3.14. First, recall that Nis also the midpoint of the segment that joins
the midpoints XandYof the diagonals ACandBD. The circle of diameter OM
goes through XandYsinceOXandOYare perpendiculars to the corresponding
diagonals, and ONis a median of the triangle OXY .
Solution 3.15. The inequality on the right-hand side follows from wx+xy+yz+
zw=(w+y)(x+z)=−(w+y)2≤0.

4.3 Solutions to the problems in Chapter 3 167
For the inequality on the left-hand side, note that
|wx+xy+yz+zw|=|(w+y)(x+z)|
≤1
2/bracketleftBig
(w+y)2+(x+z)2/bracketrightBig
≤w2+x2+y2+z2=1.
We can again use the Cauchy-Schwarz inequality to obtain
|wx+xy+yz+zw|2≤(w2+x2+y2+z2)(x2+y2+z2+w2)=1.
Solution 3.16. For the inequality on the left-hand side, rearrange as follows:
an+a2
a1+a1+a3
a2+···+an−1+a1
an=a1
a2+a2
a1+a3
a2+a2
a3+···+a1
an+an
a1,
now, use the fact that/parenleftBig
x
y+y
x/parenrightBig
≥2.
SetSn=an+a2
a1+a1+a3
a2+a2+a4
a3+···+an−1+a1
an. Using induction, prove that
Sn≤3n.
First, for n=3 ,w en e e dt os e et h a tb+c
a+c+a
b+a+b
c≤9. Ifa=b=c,
thenb+c
a+c+a
b+a+b
c= 6 and the inequality is true. Suppose that a≤b≤cand
that not all numbers are equal, then we have three cases: a=b<c,a<b= c,
a<b<c .I na l lo ft h e m ,w eh a v e a≤banda<c. Hence 2 c=c+c>a +band
a+b
c<2, and sincea+b
cis a positive integer we have c=a+b.
Thus,b+c
a+c+a
b+a+b
c=a+2b
a+2a+b
b+1=3+2b
a+2a
b.S i n c e2b
aand 2a
b
are positive integers, and since/parenleftbig
2b
a/parenrightbig/parenleftbig
2a
b/parenrightbig
= 4, we have that either both numbers
are equal to 2 or one number is 1 and the other is 4. This means the sum is at
m o s t8 ,w h i c hi sl e s st h a n9 ,t h e nt h er e s u l t .
We continue with the induction. Suppose that Sn−1≤3(n−1). Consider
{a1,…,a n}, if all are equal, then Sn=2nand the inequality is true. Suppose
instead that there are at least two diﬀerents ai’s. Take the maximum of the ai’s;
its neighbors ( ai−1,ai+1) can be equal to this maximum value, but since there are
two diﬀerent numbers between the ai’s for some maximum ai, we have that one of
its neighbors is less than ai. We can then assume, without loss of generality, that
anis maximum and that one of its neighbors, an−1ora1,i sl e s st h a n an. Then,
since 2 an>an−1+a1,w eh a v et h a tan−1+a1
an<2a n dt h e nan−1+a1
an=1 ,f o rw h i c h
an=an−1+a1. When we substitute this value of aninSn,w eg e t
Sn=an−1+a1+a2
a1+a2+a3
a2+···+an−2+an−1+a1
an−1+an−1+a1
an−1+a1
=1+an−1+a2
a1+a2+a3
a2+···+an−2+a1
an−1+1+1 .
Since Sn−1≤3(n−1), this implies that Sn≤3n.

168 Solutions to Exercises and Problems
Solution 3.17. Since the quadrilateral OBDC is cyclic, use Ptolemy’s theorem to
prove that OD=R/parenleftbigBD
BC+DC
BC/parenrightbig
,w h e r e Ris the circumradius of ABC.O nt h e
other hand, since the triangles BCE andDCA are similar, as well as the triangles
ABD andFBC, it happens that R/parenleftbigBD
BC+DC
BC/parenrightbig
=R/parenleftbigAD
FC+AD
EB/parenrightbig
. We can ﬁnd
similar equalities for OEandOF,OE=R/parenleftbigBE
AD+BE
CF/parenrightbig
andOF=R/parenleftbigCF
BE+CF
AD/parenrightbig
.
Multiplying these equalities and applying the AM-GM inequality, the result is
attained.
Another way to prove this is using inversion. Let D/prime,E/primeandF/primebe the
intersection points of AO,BOandCOwith the sides BC,CAandAB,r e –
spectively. Invert the sides BC,CAandABwith respect to ( O,R), obtain-
ing the circumcircles of the triangles OBC,OCA andOAB, respectively. Then,
OD·OD/prime=OE·OE/prime=OF·OF/prime=R2.I fx=(ABO),y=(BCO)a n d
z=(CAO), we can deduce that
AO
OD/prime=z+x
y,BO
OE/prime=x+y
zandCO
OF/prime=y+z
x.
This implies, using the AM-GM inequality, thatR3
OD/prime·OE/prime·OF/prime≥8; therefore, OD·
OE·OF≥8R3.
Solution 3.18. First, observe that AY≤2Rand that ha≤AX,w h e r e hais the
length of the altitude on BC. Then we can deduce that
/summationdisplay la
sin2A=/summationdisplay AX
AYsin2A
≥/summationdisplay ha
2Rsin2A
=/summationdisplay ha
asinA/parenleftbigg
sincesinA
a=1
2R/parenrightbigg
≥33/radicalbigg
ha
asinAhb
bsinBhc
csinC
=3
sinceha=bsinC,hb=csinA,hc=asinB.
Solution 3.19. Without loss of generality, x1≤x2≤···≤ xn.Since 1 <2<···<
n, we have, using the rearrangement inequality (1.2), that
A=x1+2×2+···+nxn≥nx 1+(n−1)x2+···+xn=B.
Then, |A+B|=|(n+1 )(x1+···+xn)|=n+ 1, hence A+B=±(n+1 ) .N o w ,
ifA+B=(n+ 1) it follows that B≤n+1
2≤A,a n di f A+B=−(n+ 1), it is
t h ec a s et h a t B≤−n+1
2≤A
If we now assume thatn+1
2or−n+1
2is between BandA,o t h e r w i s e AorB
w o u l db ei nt h ei n t e r v a l/bracketleftbig
−n+1
2,n+1
2/bracketrightbig
, then either |A|or|B|is less than or equal
ton+1
2and we can solve the problem.

4.3 Solutions to the problems in Chapter 3 169
Suppose therefore that B≤−n+1
2<n+1
2≤A.
Lety1,…,ynbe a permutation of x1,…,xnsuch that 1y 1+2y2+···+nyn=
Ctakes the maximum value with C≤−n+1
2.T a k e isuch that y1≤y2≤···≤ yi
andyi>yi+1and consider
D=y1+2y2+···+iyi+1+(i+1 )yi+(i+2 )yi+2+···+nyn
D−C=iyi+1+(i+1 )yi−(iyi+(i+1 )yi+1)=yi−yi+1>0.
Since |yi|,|yi+1|≤n+1
2, we can deduce that D−C=yi−yi+1≤n+ 1; hence
D≤C+n+ 1 and therefore C<D ≤C+n+1≤n+1
2.
On the other hand, D≥−n+1
2,s i n c e Cis the maximum sum which is less
than−n+1
2.T h u s −n+1
2≤D≤n+1
2and then |D|≤n+1
2.
Solution 3.20. Among the numbers x,y,ztwo have the same sign (say xandy),
sincec=z/parenleftBig
x
y+y
x/parenrightBig
is positive, we can deduce that zis positive.
Note that a+b−c=2xy
z,b+c−a=2yz
x,c+a−b=2zx
yare positive.
Conversely, if u=a+b−c,v=b+c−aandw=c+a−bare positive,
taking u=2xy
z,v=2yz
x,w=2zx
y, we can obtain a=u+w
2=x/parenleftBig
y
z+z
y/parenrightBig
,a n ds o
on.
Solution 3.21. First, prove that a centrally symmetric hexagon ABCDEF has
opposite parallel sides. Thus, ( ACE)=(BDF)=(ABCDEF )
2. Now, if we reﬂect
the triangle PQR with respect to the symmetry center of the hexagon, we get
the points P/prime,Q/prime,R/primewhich form the centrally symmetric hexagon PR/primeQP/primeRQ/prime,
inscribed in ABCDEF with area 2( PQR).
Solution 3.22. LetX=/summationtext4
i=1×3
i,Xi=X−x3i;i ti st h e ne v i d e n tt h a t X=
1
3/summationtext4
i=1Xi. Using the AM-GM inequality leads to1
3X1≥3/radicalbig
x3
2x33x34=1
x1; similar
inequalities hold for the other indexes and this implies that X≥/summationtext4
i=11
xi.
Using Tchebyshev’s inequality we obtain
x3
1+x32+x33+x34
4≥x21+x22+x23+x24
4·x1+x2+x3+x4
4.
Thanks to the AM-GM inequality we getx2
1+x22+x23+x24
4≥4/radicalbig
(x1x2x3x4)2=1 ,a n d
therefore X≥/summationtext
i=1xi.
Solution 3.23. Use the Cauchy-Schwarz inequality with u=/parenleftBig√
x−1
√
x,√
y−1
√
y,√
z−1
√
z/parenrightBig
andv=/parenleftbig√
x,√
y,√
z/parenrightbig
.
Solution 3.24. Ifα=∠ACM andβ=∠BDM ,t h e nMA·MB
MC·MD=t a n αtanβand
α+β=π
4. Now use the fact that tan αtanβtanγ≤tan3/parenleftBig
α+β+γ
3/parenrightBig
,w h e r e γ=π
4.
Another method uses the fact that the inequality is equivalent to ( MCD )≥
3√
3(MAB ) which is equivalent toh+l
h≥3√
3, where lis the length of the side of
the square and his the length of the altitude from MtoAB.F i n dt h em a x i m u m
h.

170 Solutions to Exercises and Problems
Solution 3.25. First note thatPL
AL+PM
BM+PN
CN=1 .N o w ,u s et h ef a c tt h a t AL,
BMandCNare less than a.
Solution 3.26. SincePB
PA·QC
QA≤1
4/parenleftBig
PB
PA+QC
QA/parenrightBig2
, it is suﬃcient to see thatPB
PA+
QC
QA=1 .
/BulletA
BCG
A/primeB/prime
C/primeP
Q
Draw BB/prime,CC/primeparallel to the median AA/primein such a way that B/primeand
C/primeare on PQ. The triangles APG andBPB/primeare similar, as well as AQG and
CQC/prime,t h u sPB
PA=BB/prime
AGandQC
QA=CC/prime
AG. Use this together with the fact that
AG=2GA/prime=BB/prime+CC/prime.
Solution 3.27. Let Γ be the circumcircle of ABC,a n dl e t Rbe its radius. Consider
the inversion in Γ. For any point Pother than O,l e tP/primebe its inverse. The inverse
of the circumcircle of OBC is the line BC,t h e n A/prime
1,t h ei n v e r s eo f A1,i st h e
intersection point between the ray OA 1andBC.S i n c e22
P/primeQ/prime=R2·PQ
OP·OQ
for two points P,Q(distinct from O)w i t hi n v e r s e s P/prime,Q/prime,w eh a v e
AA 1
OA 1=R2·A/primeA/prime
1
OA/prime·OA/prime
1·OA 1=AA/prime
1
OA=x+y+z
y+z,
where x,y,zdenote the areas of the triangles OBC,OCA,OAB, respectively.
Similarly, we have that
BB 1
OB 1=x+y+z
z+xandCC 1
OC 1=x+y+z
x+y.
Thus
AA 1
OA 1+BB 1
OB 1+CC 1
OC 1=(x+y+z)/parenleftbigg1
y+z+1
z+x+1
x+y/parenrightbigg
≥9
2.
For the last inequality, see Exercise 1.44.
22See [5, page 132] or [9, page 112].

4.3 Solutions to the problems in Chapter 3 171
Solution 3.28. The area of the triangle GBC is (GBC)=(ABC )
3=a·GL
2. Therefore
GL=2(ABC )
3a. Similarly, GN=2(ABC )
3c.
In consequence,
(GNL)=GL·GNsinB
2=4(ABC)2sinB
18ac
=4(ABC)2b2
(18abc)(2R)=(ABC)2b2
(9Rabc
4R)(4R)
=(ABC)b2
9·4R2.
Similarly, ( GLM )=(ABC )c2
9·4R2and (GMN )=(ABC )a2
9·4R2. Therefore,
(LMN )
(ABC)=1
9/parenleftbigga2+b2+c2
4R2/parenrightbigg
=R2−OG2
4R2.
The inequality in the right follows easily.
For the other inequality, note that OG=1
3OH. Since the triangle is acute,
His inside the triangle and HO≤R. Therefore,
(LMN )
(ABC)=R2−1
9OH2
4R2≥R2−1
9R2
4R2=2
9>4
27.
Solution 3.29. The function f(x)=1
1+xis convex for x>0. Thus,
f(ab)+f(bc)+f(ca)
3≥f/parenleftbiggab+bc+ca
3/parenrightbigg
=3
3+ab+bc+ca
≥3
3+a2+b2+c2=1
2,
the last inequality follows from ab+bc+ca≤a2+b2+c2.
We can also begin with
1
1+ab+1
1+bc+1
1+ca≥9
3+ab+bc+ca≥9
3+a2+b2+c2=3
2.
The ﬁrst inequality follows from inequality (1.11) and the second from Exercise
1.27.
Solution 3.30. Setx=b+2c,y=c+2a,z=a+2b. The desired inequality
becomes
/parenleftbiggx
y+y
x/parenrightbigg
+/parenleftbiggy
z+z
y/parenrightbigg
+/parenleftBigz
x+x
z/parenrightBig
+3/parenleftbiggy
x+z
y+x
z/parenrightbigg
≥15,

172 Solutions to Exercises and Problems
which can be proved using the AM-GM inequality. Another way of doing it is the
following:
a
b+2c+b
c+2a+c
a+2b=a2
ab+2ca+b2
bc+2ab+c2
ca+2bc≥(a+b+c)2
3(ab+bc+ca).
The inequality follows from inequality (1.11). It remains to prove the inequality(a+b+c)
2≥3(ab+bc+ca), which is a consequence of the Cauchy-Schwarz
inequality.
Solution 3.31. Use the inequality (1.11) or use the Cauchy-Schwarz inequality with
/parenleftbigga
√
a+b,b
√
b+c,c
√
c+d,d
√
d+a/parenrightbigg
and (√
a+b,√
b+c,√
c+d,√
d+a).
Solution 3.32. Letx=b+c−a,y=c+a−bandz=a+b−c. The similarity
between the triangles ADE andABC gives us
DE
a=perimeter of ADE
perimeter of ABC=2x
a+b+c.
Thus, DE=x(y+z)
x+y+z; that is, the inequality is equivalent tox(y+z)
x+y+z≤x+y+z
4.N o w
use the AM-GM inequality.Solution 3.33. TakeFonADwithAF=BCand deﬁne E
/primeas the intersection
ofBFandAC. Using the sine law in the triangles AE/primeF,BCE/primeandBDF,w e
obtain
AE/prime
E/primeC=AFsinF
sinE/prime·sinE/prime
BCsinB=sinF
sinB=BD
FD=AE
EC,
therefore E/prime=E.
Subsequently, consider GonBDwithBG=ADandHthe intersection
point of GEwith the parallel to BCpassing through A.U s et h ef a c tt h a tt h e
triangles ECG andEAH are similar and also use Menelaus’s theorem for the
triangle CAD with transversal EFB to conclude that AH=DB. Hence, BDAH
is a parallelogram, BH=ADandBHG is isosceles with BH=BG=AD > BE .
Solution 3.34. Note that ab+bc+ca≤3abcif and only if1
a+1
b+1
c≤3. Since
(a+b+c)/parenleftbigg1
a+1
b+1
c/parenrightbigg
≥9,
we should have that ( a+b+c)≥3. Then
3(a+b+c)≤(a+b+c)2
=/parenleftBig
a3/2a−1/2+b3/2b−1/2+c3/2c−1/2/parenrightBig2
≤/parenleftbig
a3+b3+c3/parenrightbig/parenleftbigg1
a+1
b+1
c/parenrightbigg
≤3/parenleftbig
a3+b3+c3/parenrightbig
.

4.3 Solutions to the problems in Chapter 3 173
Solution 3.35. Takeyi=xi
n−1for all i=1,2,…,n and suppose that the inequality
is false, that is,
1
1+y1+1
1+y2+···+1
1+yn>n−1.
Then
1
1+yi>/summationdisplay
j/negationslash=i/parenleftbigg
1−1
1+yj/parenrightbigg
=/summationdisplay
j/negationslash=iyj
1+yj
≥(n−1)n−1/radicalBigg
y1···ˆyi···yn
(1 +y1)···/hatwider(1 +yi)···(1 +yn),
where y1···ˆyi···ynis the product of the y’s except yi.T h e n
n/productdisplay
i=11
1+yi>(n−1)n y1···yn
(1 +y1)···(1 +yn),
a n dt h i si m p l i e s1 >x 1···xn, a contradiction.
Solution 3.36. Use the Cauchy-Schwarz inequality with/parenleftBig/radicalBig
x1
y1,…,/radicalBig
xn
yn/parenrightBig
and
/parenleftbig√
x1y1,…,√
xnyn/parenrightbig
to get
(x1+···+xn)2=/parenleftbigg/radicalbigg
x1
y1√
x1y1+···+/radicalbigg
xn
yn√
xnyn/parenrightbigg2
≤/parenleftbiggx1
y1+···+xn
yn/parenrightbigg
(x1y1+···+xnyn).
Now use the hypothesis/summationtextxiyi≤/summationtextxi.
Solution 3.37. Since abc=1 ,w eh a v e
(a−1)(b−1)(c−1) =a+b+c−/parenleftbigg1
a+1
b+1
c/parenrightbigg
and similarly
(an−1)(bn−1)(cn−1) =an+bn+cn−/parenleftbigg1
an+1
bn+1
cn/parenrightbigg
.
The proof follows from the fact that the left sides of the equalities have the same
sign.
Solution 3.38. We prove the claim using induction on n.T h ec a s e n= 1 is clear.
Now assuming the claim is true for n,w ec a np r o v ei ti st r u ef o r n+1 .
Since n<√
n2+i<n+1 ,f o r i=1,2,…,2n,w eh a v e
/braceleftBig/radicalbig
n2+i/bracerightBig
=/radicalbig
n2+i−n</radicalBigg
n2+i+/parenleftbiggi
2n/parenrightbigg2
−n=i
2n.

174 Solutions to Exercises and Problems
Thus
(n+1)2/summationdisplay
j=1/braceleftBig/radicalbig
j/bracerightBig
=n2/summationdisplay
j=1/braceleftBig/radicalbig
j/bracerightBig
+(n+1)2/summationdisplay
j=n2+1/braceleftBig/radicalbig
j/bracerightBig
≤n2−1
2+1
2n2n/summationdisplay
i=1i
=(n+1 )2−1
2.
Solution 3.39. Let us prove that the converse aﬃrmation, that is, x3+y3>2,
implies that x2+y3<x3+y4. The power mean inequality/radicalBig
x2+y2
2≤3/radicalBig
x3+y3
2
implies that
x2+y2≤(x3+y3)2/3 3√
2<(x3+y3)2/3(x3+y3)1/3=x3+y3.
Then x2−x3<y3−y2≤y4−y3. The last inequality follows from the fact that
y2(y−1)2≥0.
Second solution. Since ( y−1)2≥0, we have that 2y ≤y2+1,then2 y3≤y4+y2.
Thus, x3+y3≤x3+y4+y2−y3≤x2+y2,s i n c e x3+y4≤x2+y3.
Solution 3.40. The inequality is equivalent to
(x0−x1)+1
(x0−x1)+(x1−x2)+···+(xn−1−xn)+1
(xn−1−xn)≥2n.
Solution 3.41. Sincea+3b
4≥4√
ab3,b+4c
5≥5√
bc4andc+2a
3≥3√
ca2, we can deduce
that
(a+3b)(b+4c)(c+2a)≥60a11
12b19
20c17
15.
Now prove that c2
15≥a1
12b1
20or, equivalently, that c8≥a5b3.
Solution 3.42. We have an equivalence between the following inequalities:
7(ab+bc+ca)≤2+9abc
⇔7(ab+bc+ca)(a+b+c)≤2(a+b+c)3+9abc
⇔a2b+ab2+b2c+bc2+c2a+ca2≤2(a3+b3+c3.)
For the last one use the rearrangement inequality or Tchebyshev’s inequality.
Solution 3.43. LetEbe the intersection of ACandBD. Then the triangles ABE
andDCE are similar, which implies
|AB−CD|
|AC−BD|=|AB|
|AE−EB|.
Using the triangle inequality in ABE,w eh a v e|AB|
|AE−EB|≥1 and we therefore
conclude that |AB−CD|≥|AC−BD|. Similarly, |AD−BC|≥|AC−BD|.

4.3 Solutions to the problems in Chapter 3 175
Solution 3.44. First of all, show that a1+···+aj≥j(j+1)
2nan,f o rj≤n,i nt h e
following way. First, prove that the inequality is valid for j=n,t h a ti s , a1+···+
an≥n+1
2an;u s et h ef a c tt h a t2 ( a1+···+an)=(a1+an−1)+···+(an−1+a1)+2an.
Next, prove that if bj=a1+···+aj
1+···+j,t h e n b1≥b2≥ ··· ≥ bn≥an
n(to prove by
induction that bj≥bj+1, we need to show that bj≥aj+1
j+1which, on the other
hand, follows from the ﬁrst part for n=j+1 ) .
We provide another proof of a1+···+aj≥j(j+1)
2nan, once again using
induction. It is clear that
a1≥a1,
a1+a2
2=a1
2+a1
2+a2
2≥a2
2+a2
2=a2.
Now, let us suppose that the aﬃrmation is valid for n=1,…,j ,t h a ti s ,
a1≥a1
a1+a2
2≥a2
…
a1+a2
2+···+aj
j≥aj.
Adding all the above inequalities, we obtain
ja1+(j−1)a2
2+···+aj
j≥a1+···+aj.
Adding on both sides the identity
a1+2a2
2+···+jaj
j=aj+···+a1,
we obtain
(j+1 )/parenleftbigg
a1+a2
2+···+aj
j/parenrightbigg
≥(a1+aj)+(a2+aj−1)+···+(aj+a1)≥jaj+1.
Hence
a1+a2
2+···+aj
j≥j
j+1aj+1.
Finally, addingaj+1
j+1on both sides of the inequality provides the ﬁnal step in the
induction proof.
Now,
a1+a2
2+···+an
n=1
n(a1+···+an)+n−1/summationdisplay
j=1/parenleftbigg1
j−1
j+1/parenrightbigg
(a1+···+aj)
≥1
n/parenleftbiggn(n+1 )
2nan/parenrightbigg
+n−1/summationdisplay
j=11
j(j+1 )j(j+1 )
2nan=an.

176 Solutions to Exercises and Problems
Solution 3.45.
⎛
⎝/summationdisplay
1≤i≤nxi⎞⎠4
=⎛⎝/summationdisplay
1≤i≤nx2
i+2/summationdisplay
1≤i<j≤nxixj⎞
⎠2
≥4⎛⎝/summationdisplay
1≤i≤nx2
i⎞
⎠⎛⎝2/summationdisplay
1≤i<j≤nxixj⎞⎠
=8⎛
⎝/summationdisplay
1≤i≤nx2
i⎞
⎠⎛⎝/summationdisplay
1≤i<j≤nxixj⎞⎠
=8/summationdisplay
1≤i<j≤nxixj(x2
1+···+x2n)
≥8/summationdisplay
1≤i<j≤nxixj(x2
i+x2j).
For the ﬁrst inequality apply the AM-GM inequality.
To determine when the equality occurs, note that in the last step, two of
theximust be diﬀerent from zero and the other n−2 equal to zero; also in the
step where the AM-GM inequality was used, the xiwhich are diﬀerent from zero
should in fact be equal. We can prove that in such a case the constant C=1
8is
the minimum.
Solution 3.46. Setting3√
a=xand3√
b=y, we need to prove that ( x2+y2)3≤
2(x3+y3)2forx,y>0.
Using the AM-GM inequality we have
3x4y2≤x6+x3y3+x3y3
and
3x2y4≤y6+x3y3+x3y3,
with equality if and only if x6=x3y3=y6or, equivalently, if and only if x=y.
Adding together these two inequalities and adding x6+y6to both sides, we get
x6+y6+3x2y2(x2+y2)≤2(x6+y6+2x3y3).
Equality occurs when x=y,t h a ti s ,w h e n a=b.
Solution 3.47. Denote the left-hand side of the inequality as S.S i n c e a≥b≥c
andx≥y≥z, using the rearrangement inequality we have bz+cy≤by+cz,t h e n
(by+cz)(bz+cy)≤(by+cz)2≤2((by)2+(cz)2).
Setting α=(ax)2,β=(by)2,γ=(cz)2,w eo b t a i n
a2x2
(by+cz)(bz+cy)≥a2x2
2((by)2+(cz)2)=α
2(β+γ).

4.3 Solutions to the problems in Chapter 3 177
Adding together the other two similar inequalities, we get
S≥1
2/parenleftbiggα
β+γ+β
γ+α+γ
α+β/parenrightbigg
.
Use Nesbitt’s inequality to conclude the proof.
Solution 3.48. IfXMis a median in the triangle XYZ,t h e n XM2=1
2XY2+
1
2XZ2−1
4YZ2, a result of using Stewart’s theorem. If we take ( X,Y,Z,M )t o
be equal to ( A, B, C, P ), (B,C,D,Q ), (C,D,A,R )a n d( D,A,B,S ), and then
substitute them in the formula, we then add together the four resulting equations
to get a ﬁfth equation. Multiplying both sides of the ﬁfth equation by 4, we ﬁnd
that the left-hand side of the desired inequality equals AB2+BC2+CD2+DA2+
4(AC2+BD2). Thus, it is suﬃcient to prove that AC2+BD2≤AB2+BC2+
CD2+DA2. This inequality is known as the “parallelogram inequality”. To prove
it, let Obe an arbitrary point on the plane, and for each point Xletxdenote the
vector from OtoX. We expand each term in AB2+BC2+CD2+DA2−AC2−
BD2, writing for instance
AB2=|a−b|2=|a|2−2a·b+|b|2
and then ﬁnding that the expression equals
|a|2+|b|2+|c|2+|d|2−2(a·b+b·c+c·d+d·a−a·c−b·d)
=|a+c−b−d|2≥0,
with equality if and only if a+c=b+d, that is, only if the quadrilateral ABCD
is a parallelogram.
Solution 3.49. PutA=x2+y2+z2,B=xy+yz+zx,C=x2y2+y2z2+z2x2,
D=xyz.T h e n1= A+2B,B2=C+2xyz(x+y+z)=C+2Dandx4+y4+z4=
A2−2C=4B2−4B+1−2C=2C−4B+8D+ 1. Then, the expression in the
middle is equal to
3−2A+( 2C−4B+8D+1 )=2+2 C+8D≥2,
with equality if and only if two out of the x,y,zare zero.
Now, the right-hand expression is equal to 2+ B+D.T h u sw eh a v et op r o v e
that 2 C+8D≤B+DorB−2B2−3D≥0. Using the Cauchy-Schwarz inequality,
we get A≥B,s ot h a t B(1−2B)=BA≥B2. Thus it is suﬃcient to prove that
B2−3D=C−D≥0. But C≥xyyz+yzzx+zxxy=Das can be deduced from
the Cauchy-Schwarz inequality.Solution 3.50. Suppose that a=
x
y,b=y
z,c=z
x. The inequality is equivalent to
/parenleftbiggx
y−1+z
y/parenrightbigg/parenleftBigy
z−1+x
z/parenrightBig/parenleftBigz
x−1+y
x/parenrightBig
≤1

178 Solutions to Exercises and Problems
and can be rewritten as ( x+z−y)(x+y−z)(y+z−x)≤xyz. This last inequality
is valid if x,y,zare the lengths of the sides of a triangle. See Example 2.2.3.
A case remains when some out of the u=x+z−y,v=x+y−z,w=y+z−x
are negative. If one or three of them are negative, then the left side is negative
and the inequality is clear. If two of the values u,v,ware negative, for instance
uandv,t h e n u+v=2xis also negative; but x>0, so that this last situation is
not possible.
Solution 3.51. First note that abc≤a+b+cimplies ( abc)2≤(a+b+c)2≤
3(a2+b2+c2), where the last inequality follows from inequality (1.11).
By the AM-GM inequality, a2+b2+c2≥33/radicalbig
(abc)2,t h e n( a2+b2+c2)3≥
33(abc)2. Therefore ( a2+b2+c2)4≥32(abc)4.
Solution 3.52. Using the AM-GM inequality,
(a+b)(a+c)=a(a+b+c)+bc≥2/radicalbig
abc(a+b+c).
Second solution. Setting x=a+b,y=a+c,z=b+c, and since a,b,care
positive, we can deduce that x,y,zare the side lengths of a triangle XYZ.T h u s ,
the inequality is equivalent toxy
2≥(XYZ) as can be seen using the formula for
the area of a triangle in Section 2.2. Now, recall that the area of a triangle with
side lengths x,y,zis less than or equal toxy
2.
Solution 3.53. Since xi≥0, then xi−1≥−1. Next, we can use Bernoulli’s
inequality for all ito get
(1 + ( xi−1))i≥1+i(xi−1).
Adding these inequalities together for 1 ≤i≤n, gives us the result.
Solution 3.54. Subtracting 2, we ﬁnd that the inequalities are equivalent to
0<(a+b−c)(a−b+c)(−a+b+c)
abc≤1.
The left-hand side inequality is now obvious. The right-hand side inequality is
Example 2.2.3.
Solution 3.55. If we prove thata
√
a2+8bc≥a4/3
a4/3+b4/3+c4/3, it will be clear how to
get the result. The last inequality is equivalent to
/parenleftBig
a4/3+b4/3+c4/3/parenrightBig2
≥a2/3(a2+8bc).
Apply the AM-GM inequality to each factor of
/parenleftBig
a4/3+b4/3+c4/3/parenrightBig2
−/parenleftBig
a4/3/parenrightBig2
=/parenleftBig
b4/3+c4/3/parenrightBig/parenleftBig
a4/3+a4/3+b4/3+c4/3/parenrightBig
.

4.3 Solutions to the problems in Chapter 3 179
Another method for solving this exercise is to consider the function f(x)=1
√
x,
this function is convex for x>0(f/prime/prime(x)=3
4√
x5>0). For 0 <a,b,c<1, with
a+b+c= 1, we can deduce thata
√
x+b
√
y+c
√
z≥1
√
ax+by+cz.Applying this
tox=a2+8bc,y=b2+8caandz=c2+8ab(previously multiplying by an
appropriate factor to have the condition a+b+c=1 ) ,w eg e t
a
√
a2+8bc+b
√
b2+8ca+c
√
c2+8ab≥1
√
a3+b3+c3+2 4abc.
Also use the fact that
(a+b+c)3=a3+b3+c3+3 (a2b+a2c+b2a+b2c+c2a+c2b)+6abc
≥a3+b3+c3+2 4abc.
Solution 3.56. Using the Cauchy-Schwarz inequality/summationtextaibi≤/radicalbig
/summationtexta2
i/radicalbig
/summationtextb2iwith
ai=1 ,bi=xi
1+x2
1+x22+···+x2
i, we can deduce that
x1
1+x2
1+x2
1+x21+x22+···+xn
1+x21+···+x2n≤√
n/radicalBig
/summationdisplay
b2
i.
Then, it suﬃces to show that/summationtextb2i<1.
Observe that for i≥2,
b2
i=/parenleftbiggxi
1+x2
1+···+x2
i/parenrightbigg2
=x2i
(1 +x2
1+···+x2
i)2
≤x2i
(1 +x2
1+···+x2
i−1)(1 + x2
1+···+x2
i)
=1
(1 +x2
1+···+x2
i−1)−1
(1 +x2
1+···+x2
i).
Fori=1 ,u s et h ef a c tt h a t b21≤x2
1
1+x2
1=1−1
1+x21. Adding together these inequal-
ities, the right-hand side telescopes to yield
/summationdisplay
b2
i=n/summationdisplay
i=1/parenleftbiggxi
1+x2
1+···+x2
i/parenrightbigg2
≤1−1
1+x2
1+···+x2n<1.
Solution 3.57. Since there are only two possible values for α,β,γ, the three must
either all be equal, or else two are equal and one is diﬀerent from these two.
Therefore, we have two cases to consider.
(1)α=β=γ.I nt h i sc a s ew eh a v e a+b+c= 0, and therefore
/parenleftbigga3+b3+c3
abc/parenrightbigg2
=/parenleftbigga3+b3−(a+b)3
−ab(a+b)/parenrightbigg2
=/parenleftbigg(a+b)2−a2+ab−b2
ab/parenrightbigg2
=/parenleftbigg3ab
ab/parenrightbigg2
=9.

180 Solutions to Exercises and Problems
(2) Without loss of generality, we assume that α=β,γ/negationslash=α,t h e n c=a+b
and
a3+b3+c3
abc=a3+b3+(a+b)3
ab(a+b)=(a+b)2+a2−ab+b2
ab
=2a2+2b2+ab
ab=2/parenleftbigga
b+b
a/parenrightbigg
+1.
Ifaandbhave the same sign, we see that this expression is not less than 5, and
its square is therefore no less than 25. If the signs of aandbare not the same, we
havea
b+b
a≤−2, therefore 2/parenleftbiga
b+b
a/parenrightbig
+1≤−3a n d/parenleftbig
2/parenleftbiga
b+b
a/parenrightbig
+1/parenrightbig2≥9.
Thus, the smallest possible value is 9.
Solution 3.58. Using the AM-GM inequality,1
b(a+b)+1
c(b+c)+1
a(c+a)≥3
XY,w h e r e
X=3√
abc,Y=3/radicalbig
(a+b)(b+c)(c+a). Using AM-GM inequality again gives
X≤a+b+c
3andY≤2a+b+c
3,t h e n3
XY≥/parenleftbig27
2/parenrightbig1
(a+b+c)2.
Solution 3.59. The inequality is equivalent to a4+b4+c4≥a2bc+b2ca+c2ab,
which follows using Muirhead’s theorem since [4 ,0,0]≥[2,1,1].
Second solution.
a3
bc+b3
ca+c3
ab=a4
abc+b4
abc+c4
abc
≥(a2+b2+c2)2
3abc
≥(a+b+c)4
27abc=/parenleftbigga+b+c
3/parenrightbigg3(a+b+c)
abc
≥(abc)/parenleftbigga+b+c
abc/parenrightbigg
=a+b+c.
In the ﬁrst two inequalities we applied inequality (1.11), and in the last inequality
we used the AM-GM inequality.
Solution 3.60. Takef(x)a sf(x)=x
1−x.S i n c e f/prime/prime(x)=2
(1−x)3>0,f(x)i sc o n v e x .
Using Jensen’s inequality we get f(x)+f(y)+f(z)≥3f(x+y+z
3). But since fis
increasing for x<1, and because the AM-GM inequality helps us to establish
thatx+y+z
3≥3√
xyz, then we can deduce that f(x+y+z
3)≥f(3√
xyz).
Solution 3.61.
/parenleftbigga
b+c+1
2/parenrightbigg/parenleftbiggb
c+a+1
2/parenrightbigg/parenleftbiggc
a+b+1
2/parenrightbigg
≥1
is equivalent to (2 a+b+c)(2b+c+a)(2c+a+b)≥8(b+c)(c+a)(a+b). Now,
observe that (2 a+b+c)=(a+b+a+c)≥2/radicalbig
(a+b)(c+a).

4.3 Solutions to the problems in Chapter 3 181
Solution 3.62. The inequality of the problem is equivalent to the following inequal-
ity:
(a+b−c)(a+b+c)
c2+(b+c−a)(b+c+a)
a2+(c+a−b)(c+a+b)
b2≥9,
w h i c hi nt u r ni se q u i v a l e n tt o(a+b)2
c2+(b+c)2
a2+(c+a)2
b2≥12. Since ( a+b)2≥4ab,
(b+c)2≥4bcand (c+a)2≥4ca, we can deduce that
(a+b)2
c2+(b+c)2
a2+(c+a)2
b2≥4ab
c2+4bc
a2+4ca
b2≥123/radicalbigg
(ab)(bc)(ca)
c2a2b2=1 2.
Solution 3.63. By the AM-GM inequality, x2+√
x+√
x≥3x. Adding similar
inequalities for y,z,w eg e t x2+y2+z2+2 (√
x+√
y+√
z)≥3(x+y+z)=
(x+y+z)2=x2+y2+z2+2 (xy+yz+zx).
Solution 3.64. If we multiply the equality 1 =1
a+1
b+1
cby√
abc,w eg e t√
abc=/radicalBig
ab
c+/radicalBig
bc
a+/radicalbig
ca
b. Then, it is suﬃcient to prove that√
c+ab≥√
c+/radicalBig
ab
c.
Squaring shows that this is equivalent to c+ab≥c+ab
c+2√
ab,c+ab≥
c+ab(1−1
a−1
b)+2√
abora+b≥2√
ab.
Solution 3.65. Since (1 −a)(1−b)(1−c)=1−(a+b+c)+ab+bc+ca−abcand
sincea+b+c= 2, the inequality is equivalent to
0≤(1−a)(1−b)(1−c)≤1
27.
Buta<b+c=2−aimplies that a<1 and, similarly, b<1a n d c<1, therefore
the left inequality is true. The other one follows from the AM-GM inequality.
Solution 3.66. It is possible to construct another triangle AA 1Mwith sides AA 1,
A1M,MAof lengths equal to the lengths of the medians ma,mb,mc.
A
CC1
A1B1M
B
Moreover, ( AA 1M)=3
4(ABC). Then the inequality we have to prove is
1
mamb+1
mbmc+1
mcma≤3
4√
3
(AA 1M).

182 Solutions to Exercises and Problems
Now, the last inequality will be true if the triangle with side-lengths a,b,cand
areaSsatisﬁes the following inequality:
1
ab+1
bc+1
ca≤3√
3
4S.
Or equivalently, 4√
3S≤9abc
a+b+c, which is Example 2.4.6.
Solution 3.67. Substitute cd=1
abandda=1
bc, so that the left-hand side (LHS)
inequality becomes
1+ab
1+a+1+ab
ab+abc+1+bc
1+b+1+bc
bc+bcd(4.8)
=( 1+ ab)/parenleftbigg1
1+a+1
ab+abc/parenrightbigg
+( 1+ bc)/parenleftbigg1
1+b+1
bc+bcd/parenrightbigg
.
Now, using the inequality1
x+1
y≥4
x+y,w eg e t
(LHS) ≥(1 +ab)4
1+a+ab+abc+( 1+ bc)4
1+b+bc+bcd
=4/parenleftbigg1+ab
1+a+ab+abc+1+bc
1+b+bc+bcd/parenrightbigg
=4/parenleftbigg1+ab
1+a+ab+abc+a+abc
a+ab+abc+abcd/parenrightbigg
=4.
Solution 3.68. Using Stewart’s theorem we can deduce that
l2
a=bc/parenleftBigg
1−/parenleftbigga
b+c/parenrightbigg2/parenrightBigg
=bc
(b+c)2((b+c)2−a2)≤1
4((b+c)2−a2).
Using the Cauchy-Schwarz inequality leads us to
(la+lb+lc)2≤3(l2
a+l2
b+l2
c)
≤3
4((a+b)2+(b+c)2+(c+a)2−a2−b2−c2)
≤3
4(a+b+c)2.
Solution 3.69. Since1
1−a=1
b+c, the inequality is equivalent to
1
b+c+1
c+a+1
a+b≥2
2a+b+c+2
2b+a+c+2
2c+a+b.
Now, using the fact that1
x+1
y≥4
x+y,w eh a v e
2/parenleftbigg1
b+c+1
c+a+1
a+b/parenrightbigg
≥4
a+b+2c+4
b+c+2a+4
c+a+2b
which proves the inequality.

4.3 Solutions to the problems in Chapter 3 183
Solution 3.70. We may take a≤b≤c.T h e n c<a +band
n√
2
2=n√
2
2(a+b+c)>n√
2
2(2c)=n√
2cn≥n√
bn+cn.
Since a≤b, we can deduce that
/parenleftBig
b+a
2/parenrightBign
=bn+nbn−1a
2+ other positive terms
>bn+n
2abn−1≥bn+an.
Similarly, since a≤c,w eh a v e( c+a
2)n>cn+an, therefore
(an+bn)1
n+(bn+cn)1
n+(cn+an)1
n<b+a
2+n√
2
2+c+a
2
=a+b+c+n√
2
2=1+n√
2
2.
Second solution. Remember that a,b,care the lengths of the sides of a triangle if
and only if there exist positive numbers x,y,zwitha=y+z,b=z+x,c=x+y.
Since a+b+c= 1, we can deduce that x+y+z=1
2.
Now, we use Minkowski’s inequality
/parenleftBiggn/summationdisplay
i=1(xi+yi)m/parenrightBigg 1
m
≤/parenleftBiggn/summationdisplay
i=1xm
i/parenrightBigg 1
m
+/parenleftBiggn/summationdisplay
i=1ym
i/parenrightBigg 1
m
to get
(an+bn)1
n=( (y+z)n+(z+x)n)1
n≤(xn+yn)1
n+( 2zn)1
n<c+n√
2z.
Similarly, ( bn+cn)1
n<a+n√
2xand (cn+an)1
n<b+n√
2y. Therefore
(an+bn)1
n+(bn+cn)1
n+(cn+an)1
n<a+b+c+n√
2(x+y+z)=1+n√
2
2.
Solution 3.71. F i r s tn o t i c et h a ti fw er e s t r i c tt h es u m st o i<j, then they are
halved. The left-hand side sum is squared while the right-hand side sum is not,
so that the desired inequality with sums restricted to i<j has (1 /3), instead of
(2/3), on the right-hand side.
Consider the sum of all the |xi−xj|withi<j.T h en u m b e r x1appears in
(n−1) terms with negative sign, x2appears in one term with positive sign and
(n−2) terms with negative sign, and so on. Thus, we get
−(n−1)x1−(n−3)x2−(n−5)x3−···+(n−1)xn=/summationdisplay
(2i−1−n)xi.

184 Solutions to Exercises and Problems
We can now apply the Cauchy-Schwarz inequality to show that the square of this
sum is less than/summationtextx2
i/summationtext(2i−1−n)2.
Looking at the sum at the other side of the desired inequality, we immediately
s e et h a ti ti s n/summationtextx2i−(/summationtextxi)2. We would like to get rid of the second term, which
is easy because if we add hto every xithe sums in the desired inequality are
unaﬀected (since they only involve diﬀerences of the xi), so we can choose an hto
make/summationtextxizero. Thus, we can ﬁnish if we can prove that/summationtext(2i−1−n)2=n(n2−1)
3,
/summationdisplay
(2i−1−n)2=4/summationdisplay
i2−4(n+1 )/summationdisplay
i+n(n+1 )2
=2
3n(n+ 1)(2 n+1 )−2n(n+1 )2+n(n+1 )2
=1
3n(n+ 1)(2(2 n+1 )−6(n+1 )+3 ( n+1 ) )
=1
3n(n2−1).
This establishes the required inequality.
Second solution. The inequality is of the Cauchy-Schwarz type, and since the prob-
lem asks us to prove that equality holds when x1,x2,…,xnform an arithmetic
progression, that is, when xi−xj=r(i−j)w i t hr> 0, then consider the following
inequality which is true, as can be inferred from the Cauchy-Schwarz inequality,
⎛
⎝/summationdisplay
i,j|i−j||xi−xj|⎞⎠2
≤/summationdisplay
i,j(i−j)2/summationdisplay
i,j(xi−xj)2.
Here, we already know that equality holds if and only if (x i−xj)=r(i−j), with
r>0.
Since/summationtext
i,j(i−j)2=( 2n−2)·12+( 2n−4)·22+···+2·(n−1)2=n2(n2−1)
6,
we need to prove that/summationtext
i,j|i−j||xi−xj|=n
2/summationtext
i,j|xj−xj|. To see that it happens
compare the coeﬃcient of xiin each side. On the left-hand side the coeﬃcient is
(i−1) + (i−2) +···+(i−(i−1))−((i+1 )−i)+( (i+2 )−i)+···+(n−i))
=(i−1)i
2−(n−i)(n−i+1 )
2=n(2i−n−1)
2.
The coeﬃcient of xion the right-hand side is
n
2⎛⎝/summationdisplay
i<j1+/summationdisplay
j>i−1⎞⎠=n
2((i−1)−(n−i)) =n(2i−n−1)
2.
Since they are equal we have ﬁnished the proof.

4.3 Solutions to the problems in Chapter 3 185
Solution 3.72. Letxn+1=x1andxn+2=x2. Deﬁne
ai=xi
xi+1and bi=xi+xi+1+xi+2,i∈{1,…,n }.
It is evident that
n/productdisplay
i=1ai=1 ,n/summationdisplay
i=1bi=3n/summationdisplay
i=1xi=3.
The inequality is equivalent to
n/summationdisplay
i=1ai
bi≥n2
3.
Using the AM-GM inequality, we can deduce that
1
nn/summationdisplay
i=1bi≥n/radicalbig
b1···bn⇔3
n≥n/radicalbig
b1···bn⇔1
n√
b1···bn≥n
3.
On the other hand and using again the AM-GM inequality, we get
n/summationdisplay
i=1ai
bi≥nn/radicalbigg
a1
b1···an
bn=nn√
a1···an
n√
b1···bn=n
n√
b1···bn≥n2
3.
Solution 3.73. For any apositive real number, a+1
a≥2, with equality occurring
if and only if a=1 .S i n c et h en u m b e r s ab,bcandcaare non-negative, we have
P(x)P/parenleftbigg1
x/parenrightbigg
=(ax2+bx+c)/parenleftbigg
a1
x2+b1
x+c/parenrightbigg
=a2+b2+c2+ab/parenleftbigg
x+1
x/parenrightbigg
+bc/parenleftbigg
x+1
x/parenrightbigg
+ca/parenleftbigg
x2+1
x2/parenrightbigg
≥a2+b2+c2+2ab+2bc+2ca=(a+b+c)2=P(1)2.
Equality takes place if and only if either x=1o r ab=bc=ca=0 ,w h i c hi nv i e w
of the condition a>0m e a n st h a t b=c= 0. Consequently, for any positive real
xwe have
P(x)P/parenleftbigg1
x/parenrightbigg
≥(P(1))2
with equality if and only if either x=1o r b=c=0 .

186 Solutions to Exercises and Problems
Second solution. Using the Cauchy-Schwarz inequality we get
P(x)P/parenleftbigg1
x/parenrightbigg
=(ax2+bx+c)/parenleftbigg
a1
x2+b1
x+c/parenrightbigg
=/parenleftBig
(√
ax)2+(√
bx)2+(√
c)2/parenrightBig⎛
⎝/parenleftbigg√
a
x/parenrightbigg2
+/parenleftBigg√
b
√
x/parenrightBigg2
+(√
c)2⎞⎠
≥/parenleftBigg
√
ax√
a
x+√
bx√
b
√
x+√
c√
c/parenrightBigg2
=(a+b+c)2=(P(1))2.
Solution 3.74.
a2(b+c)+b2(c+a)+c2(a+b)
(a+b)(b+c)(c+a)≥3
4
⇔a2b+a2c+b2c+b2a+c2a+c2b
2abc+a2b+a2c+b2c+b2a+c2a+c2b≥3
4
⇔a2b+a2c+b2c+b2a+c2a+c2b−6abc≥0
⇔[2,1,0]≥[1,1,1].
The last inequality follows after using Muirhead’s theorem.
Second solution. Use inequality (1.11) and the Cauchy-Schwarz inequality.
Solution 3.75. Applying the AM-GM inequality to each denominator, one obtains
1
1+2ab+1
1+2bc+1
1+2ca≥1
1+a2+b2+1
1+b2+c2+1
1+c2+a2.
Now, using inequality (1.11) leads us to
1
1+a2+b2+1
1+b2+c2+1
1+c2+a2≥( 1+1+1 )2
3+2 ( a2+b2+c2)=9
3+2· 3=1.
Solution 3.76. The inequality is equivalent to each of the following ones:
x4+y4+z4+3 (x+y+z)≥−(x3z+x3y+y3x+y3z+z3y+z3x),
x3(x+y+z)+y3(x+y+z)+z3(x+y+z)+3 (x+y+z)≥0,
(x+y+z)(x3+y3+z3−3xyz)≥0.
Identity (1.9) shows us that the last inequality is equivalent to
1
2(x+y+z)2((x−y)2+(y−z)2+(z−x)2)≥0.

4.3 Solutions to the problems in Chapter 3 187
Solution 3.77. LetOandIbe the circumcenter and the incenter of the acute
triangle ABC, respectively. The points O,M,Xare collinear and OCX and
OMC are similar right triangles. Hence we have
OC
OX=OM
OC.
Since OC=R=OA,w eh a v eOA
OM=OX
OA. Hence OAM andOXA are similar, so
we haveAM
AX=OM
R.
It now suﬃces to show that OM≤r. Let us compare the angles ∠OBM and
∠IBM.S i n c eABC is an acute triangle, OandIlie inside ABC.N o ww eh a v e
∠OBM =π
2−∠A=1
2(∠A+∠B+∠C)−∠A=1
2(∠B+∠C−∠A)≤∠B
2=∠IBM.
Similarly, we have ∠OCM ≤∠ICM.T h u st h ep o i n t Olies inside IBC,s ow eg e t
OM≤r.
Solution 3.78. Setting a=x2,b=y2,c=z2, the inequality is equivalent to
x6+y6+z6≥x4yz+y4zx+z4xy.
This follows from Muirhead’s theorem since [6 ,0,0]≥[4,1,1].
Solution 3.79. Use the Cauchy-Schwarz inequality to see that√
xy+z=√
x√
y+√
z√
z≤√
x+z√
y+z=/radicalbig
xy+z(x+y+z)=√
xy+z. Similarly,√
yz+x≤√
yz+xand√
zx+y≤√
zx+y. Therefore,
√
xy+z+√
yz+x+√
zx+y≥√
xy+√
yz+√
zx+x+y+z.
Solution 3.80. Using Example 1.4.11, we have
a3+b3+c3≥(a+b+c)(a2+b2+c2)
3.
Now,
a3+b3+c3≥a+b+c
3(a2+b2+c2)≥3√
abc(ab+bc+ca)≥ab+bc+ca,
where we have used the AM-GM and the Cauchy-Schwarz inequalities.
Solution 3.81. Using Example 1.4.11, we get ( a+b+c)(a2+b2+c2)≤3(a3+b3+c3),
but by hypothesis a2+b2+c2≥3(a3+b3+c3), hence a+b+c≤1. On the other
hand,
4(ab+bc+ca)−1≥a2+b2+c2≥ab+bc+ca,
therefore 3( ab+bc+ca)≥1. As
1≤3(ab+bc+ca)≤(a+b+c)2≤1,
we obtain a+b+c=1 .C o n s e q u e n t l y , a+b+c=1a n d3 ( ab+bc+ca)=( a+b+c)2,
which implies a=b=c=1
3.

188 Solutions to Exercises and Problems
Solution 3.82. The Cauchy-Schwarz inequality yields
(|a|+|b|+|c|)2≤3(a2+b2+c2)=9.
Hence |a|+|b|+|c|≤3. From the AM-GM inequality it follows that
a2+b2+c2≥33/radicalbig
(abc)2
or|abc|≤1, which implies −abc≤1. The requested inequality is then obtained
by summation.
Solution 3.83. Notice that
OA 1
AA 1=(OBC)
(ABC)=OB·OC·BC
4R1·4R
AB·AC·BC.
Now, we have to prove that
OB·OC·BC+OA·OB·AB+OA·OC·AC≥AB·AC·BC.
We consider the complex coordinates O(0),A(a),B(b),C(c)a n do b t a i n
|b|·|c|·|b−c|+|a|·|b|·|a−b|+|a|·|c|·|c−a|≥|a−b|·|b−c|·|c−a|.
That is,
|b2c−c2b|+|a2b−b2a|+|c2a−a2c|≥|ab2+bc2+ca2−a2b−b2c−c2a|,
which is obvious by the triangle inequality.
Solution 3.84. LetS={i1,i1+1,…,j 1,i2,i2+1,…,j 2,…,i p,…,j p}be the
ordering of S,w h e r e jk<ik+1fork=1,2,…,p −1. Take Sp=a1+a2+···+ap,
S0=0 .T h e n
/summationdisplay
i∈Sai=Sjp−Sip−1+Sjp−1−Sip−1−1+···+Sj1−Si1−1
and /summationdisplay
1≤i≤j≤n(ai+···+aj)2=/summationdisplay
0≤i≤j≤n(Si−Sj)2.
It suﬃces to prove an inequality with the following form:
(x1−x2+···+(−1)p+1xp)2≤/summationdisplay
1≤i<j≤p(xj−xi)2+p/summationdisplay
i=1×2
i, (4.9)
because this means neglecting the same non-negative terms on the right-hand side
of the inequality. Thus inequality (4.9) reduces to
4/summationdisplay
1≤i≤j≤p
j−ievenxixj≤(p−1)p/summationdisplay
i=1×2
i.

4.3 Solutions to the problems in Chapter 3 189
This can be obtained adding together inequalities with the form 4x ixj≤2(x2
i+x2j),
i<j,j−i=even (for odd i,xiappears in such inequality/bracketleftbigp−1
2/bracketrightbig
times, and for
eveni,xiappears in such inequality/bracketleftbigp
2/bracketrightbig
−1t i m e s ) .
Solution 3.85. Letx=a+b+c,y=ab+bc+ca,z=abc.T h e n a2+b2+c2=x2−2y,
a2b2+b2c2+c2a2=y2−2xz,a2b2c2=z2, and the inequality to be proved becomes
z2+2(y2−2xz)+4(x2−2y)+8≥9yorz2+2y2−4xz+4×2−17y+8≥0. Now,
froma2+b2+c2≥ab+bc+ca=ywe obtain x2=a2+b2+c2+2y≥3y.
Also,
a2b2+b2c2+c2a2=(ab)2+(bc)2+(ca)2
≥ab·ac+bc·ab+ac·bc
=(a+b+c)abc=xz,
and thus y2=a2b2+b2c2+c2a2+2xz≥3xz. Hence,
z2+2y2−4xz+4×2−17y+8=/parenleftBig
z−x
3/parenrightBig2
+8
9(y−3)2+10
9(y2−3xz)
+35
9(x2−3y)≥0,
as required.
Second solution. Expanding the left-hand side of the inequality we obtain the
equivalent inequality
(abc)2+2 (a2b2+b2c2+c2a2)+4 ( a2+b2+c2)+8≥9(ab+bc+ca).
Since 3( a2+b2+c2)≥3(ab+bc+ca)a n d2 ( a2b2+b2c2+c2a2)+6≥4(ab+bc+ca)
(for instance, 2 a2b2+2≥4√
a2b2=4ab), it is enough to prove that
(abc)2+a2+b2+c2+2≥2(ab+bc+ca).
Part (i) of Exercise 1.90 tells us that it is enough to prove that ( abc)2+2≥
33√
a2b2c2, but this follows from the AM-GM inequality.
Solution 3.86. Let us write
3
3√
33/radicalbigg
1
abc+6 (a+b+c)=3
3√
33/radicalbigg
1+6a2bc+6b2ac+6c2ab
abc
=3
3√
33/radicalbigg
1+3ab(ac+bc)+3bc(ba+ca)+3 ca(ab+bc)
abc,
and consider the condition ab+bc+ca= 1 to obtain
3
3√
33/radicalbigg
1+3ab−3(ab)2+3bc−3(bc)2+3ca−3(ca)2
abc
=3
3√
33/radicalbigg
4−3((ab)2+(bc)2+(ca)2)
abc.

190 Solutions to Exercises and Problems
It is easy to see that 3((ab )2+(bc)2+(ac)2)≥(ab+bc+ac)2(use the Cauchy-
Schwarz inequality). Then, it is enough to prove that
3
3√
33/radicalbigg
3
abc≤1
abc,
which is equivalent to ( abc)2≤1
27. But this last inequality follows from the AM-
GM inequality,
(abc)2=(ab)(bc)(ca)≤/parenleftbiggab+bc+ca
3/parenrightbigg3
=1
27.
The equality holds if and only if a=b=c=1
√
3.
Solution 3.87. Using symmetry, it suﬃces to prove that t1<t 2+t3.W eh a v e
n/summationdisplay
i=1tin/summationdisplay
i=11
ti=n+/summationdisplay
1≤i<j≤n/parenleftbiggti
tj+tj
ti/parenrightbigg
=n+t1/parenleftbigg1
t2+1
t3/parenrightbigg
+1
t1(t2+t3)+/summationdisplay
(i,j)/negationslash=(1,2),(1,3)/parenleftbiggti
tj+tj
ti/parenrightbigg
.
Using the AM-GM inequality we get
/parenleftbigg1
t2+1
t3/parenrightbigg
≥2
√
t2t3,t2+t3≥2√
t2t3andti
tj+tj
ti≥2 for all i,j.
Thus, setting a=t1/√
t2t3>0 and using the hypothesis, we arrive at
n2+1>n/summationdisplay
i=1tin/summationdisplay
i=11
ti≥n+2t1
√
t2t3+2√
t2t3
t1+2/bracketleftbiggn2−n
2−2/bracketrightbigg
=2a+2
a+n2−4.
Hence 2 a+2
a−5<0, which implies 1 /2<a=t1/√
t2t3<2. Sot1<2√
t2t3,a n d
one more application of the AM-GM inequality yields t1<2√
t2t3≤t2+t3,a s
needed.
Solution 3.88. Note that 1 + b−c=a+b+c+b−c=a+2b≥0. Then
a3√
1+b−c≤a/parenleftbigg1+1+( 1+ b−c)
3/parenrightbigg
=a+ab−ac
3.
Similarly,
b3√
1+c−a≤b+bc−ba
3
c3√
1+a−b≤c+ca−cb
3.
Adding these three inequalities, we get
a3√
1+b−c+b3√
1+c−a+c3√
1+a−b≤a+b+c=1.

4.3 Solutions to the problems in Chapter 3 191
Solution 3.89. If any of the numbers is zero or if an odd number of them are
negative, then x1x2···x6≤0 and the inequality follows.
Therefore, it can only be 2 or 4 negative numbers between the numbers
in the inequality. Suppose that neither of them are zero and that there are 2
negative numbers (in the other case, change the signs of all numbers). If yi=|xi|,
then it is clear that y2
1+y2
2+···+y2
6=6 ,y1+y2=y3+···+y6and that
x1x2···x6=y1y2···y6.
From the AM-GM inequality we get
y1y2≤/parenleftbiggy1+y2
2/parenrightbigg2
=A2.
Also, the AM-GM inequality yields
y3y4y5y6≤/parenleftbiggy3+y4+y5+y6
4/parenrightbigg4
=/parenleftbiggy1+y2
4/parenrightbigg4
=1
24A4.
Therefore, y1y2···y6≤1
24A6.
On the other hand, the Cauchy-Schwarz inequality implies that
2(y2
1+y2
2)≥(y1+y2)2=4A2
4(y2
3+y2
4+y2
5+y2
6)≥(y3+y4+y5+y6)2=4A2.
Thus, 6 = y2
1+y2
2+···+y2
6≥2A2+A2=3A2and then y1y2···y6≤1
24A6≤
23
24=1
2.
Solution 3.90. Use the Cauchy-Schwarz inequality with (1 ,1,1) and (a
b,b
c,c
a)t o
obtain
(12+12+12)/parenleftbigga2
b2+b2
c2+c2
a2/parenrightbigg
≥/parenleftbigga
b+b
c+c
a/parenrightbigg2
.
The AM-GM inequality leads us toa
b+b
c+c
a≥33/radicalBig
abc
bca=3 ,t h e n
/parenleftbigga2
b2+b2
c2+c2
a2/parenrightbigg
≥/parenleftbigga
b+b
c+c
a/parenrightbigg
.
Similarly,a
c+b
a+c
b≥33/radicalBig
abc
bca= 3. Therefore,
a2
b2+b2
c2+c2
a2+a
c+b
a+c
b≥3+a
b+b
c+c
a.
Addinga
c+b
a+c
bto both sides yields the result.

192 Solutions to Exercises and Problems
Solution 3.91. Note that
a2+2
2=(a2−a+1 )+( a+1 )
2≥/radicalbig
(a2−a+1 ) (a+1 )=/radicalbig
1+a3.
After substituting in the given inequality, we need to prove that
a2
(a2+2 ) (b2+2 )+b2
(b2+2 ) (c2+2 )+c2
(c2+2 ) (a2+2 )≥1
3.
Setx=a2,y=b2,z=c2,t h e n xyz=6 4a n d
x
(x+2 ) (y+2 )+y
(y+2 ) (z+2 )+z
(z+2 ) (x+2 )≥1
3
if and only if
3[x(z+2 )+ y(x+2 )+ z(y+2 ) ]≥(x+2 ) (y+2 ) (z+2 ).
Now, 3( xy+yz+zx)+6 ( x+y+z)≥xyz+2 (xy+yz+zx)+4 ( x+y+z)+8
if and only if xy+yz+zx+2 (x+y+z)≥xyz+ 8 = 72, but using the AM-GM
inequality leads to x+y+z≥12 and xy+yz+zx≥48, which ﬁnishes the proof.
Solution 3.92. Observe that
x5−x2
x5+y2+z2−x5−x2
x3(x2+y2+z2)=x2(y2+z2)(x3−1)2
x3(x5+y2+z2)(x2+y2+z2)≥0.
Then
/summationdisplay x5−x2
x5+y2+z2≥/summationdisplay x5−x2
x3(x2+y2+z2)
=1
x2+y2+z2/summationdisplay/parenleftbigg
x2−1
x/parenrightbigg
≥1
x2+y2+z2/summationdisplay
(x2−yz)≥0.
The second inequality follows from the fact that xyz≥1, that is,1
x≤yz.T h e
last inequality follows from (1.8).
Second solution. First, note that
x5−x2
x5+y2+z2=x5+y2+z2−(x2+y2+z2)
x5+y2+z2=1−x2+y2+z2
x5+y2+z2.
Now we need to prove that
1
x5+y2+z2+1
x5+z2+x2+1
x5+x2+y2≤3
x2+y2+z2.

4.3 Solutions to the problems in Chapter 3 193
Using the Cauchy-Schwarz inequality we get
(x2+y2+z2)2≤(x2·x3+y2+z2)(x2·1
x3+y2+z2)
and since xyz≥1, then x2·1
x3=1
x≤yz, and we have that
(x2+y2+z2)2≤(x5+y2+z2)(yz+y2+z2)
therefore
/summationdisplay 1
x5+y2+z2≤/summationdisplayyz+y2+z2
(x2+y2+z2)2≤/summationdisplayy2+z2
2+y2+z2
(x2+y2+z2)2
=3
x2+y2+z2.
Solution 3.93. Notice that
(1 +abc)/parenleftbigg1
a(b+1 )+1
b(c+1 )+1
c(a+1 )/parenrightbigg
+3
=1+abc+ab+a
a(b+1 )+1+abc+bc+b
b(c+1 )+1+abc+ca+c
c(a+1 )
=1+a
a(b+1 )+b(c+1 )
(b+1 )+1+b
b(c+1 )+c(a+1 )
(c+1 )+1+c
c(a+1 )+a(b+1 )
(a+1 )≥6.
The last inequality follows after using the AM-GM inequality for six numbers.
Solution 3.94. LetRbe the circumradius of the triangle ABC.S i n c e ∠BOC =
2∠A,∠COA =2∠Band∠AOB =2∠C,w eh a v et h a t
(ABC)=(BOC)+(COA)+(AOB)=R2
2(sin 2A+s i n2 B+s i n2 C)
≤R2
23s i n/parenleftbigg2A+2B+2C
3/parenrightbigg
=R2
23s i n/parenleftbigg2π
3/parenrightbigg
=3√
3R2
4.
The inequality follows since the function sin xis concave in [0 ,π].
On the other hand, since BOC is isosceles, the perpendicular bisector OA/primeof
BCis also the internal bisector of the angle ∠BOC,s ot h a t ∠BOA/prime=∠COA/prime=
∠A; similarly ∠COB/prime=∠AOB/prime=∠Band∠AOC/prime=∠BOC/prime=∠C.I nt h e
triangle B/primeOC/primethe altitude on the side B/primeC/primeisR
2andB/primeC/prime=R
2(tanB+t a n C).

194 Solutions to Exercises and Problems
Therefore, the area of the triangle B/primeOC/primeis (B/primeOC/prime)=R2
8(tanB+t a n C). Simi-
larly, ( C/primeOA/prime)=R2
8(tanC+t a n A)a n d( A/primeOB/prime)=R2
8(tanA+t a n B). Then,
(A/primeB/primeC/prime)=(B/primeOC/prime)+(C/primeOA/prime)+(A/primeOB/prime)=R2
4(tanA+t a n B+t a n C)
≥R2
43t a n/parenleftbiggA+B+C
3/parenrightbigg
=R2
43t a n/parenleftBigπ
3/parenrightBig
=3√
3R2
4.
The inequality follows since the function tan xis convex in [0 ,π
2].
Hence,
(A/primeB/primeC/prime)≥3√
3R2
4≥(ABC).
Solution 3.95. First, note that a2+bc≥2√
a2bc=2√
ab√
caand similarly b2+ca≥
2√
bc√
ab,c2+ab≥2√
ca√
bc; then it follows that
1
a2+bc+1
b2+ca+1
c2+ab≤1
2/parenleftbigg1
√
ab√
ca+1
√
bc√
ab+1
√
ca√
bc/parenrightbigg
.
Now, using the Cauchy-Schwarz inequality in the following way
/parenleftbigg1
√
ab√
ca+1
√
bc√
ab+1
√
ca√
bc/parenrightbigg2
≤/parenleftbigg1
ab+1
bc+1
ca/parenrightbigg/parenleftbigg1
ca+1
ab+1
bc/parenrightbigg
,
the result follows.
Solution 3.96. From the Cauchy-Schwarz inequality we get
/summationdisplay
i/negationslash=jai
aj/summationdisplay
i/negationslash=jaiaj≥⎛
⎝/summationdisplay
i/negationslash=jai⎞⎠2
=/parenleftBigg
(n−1)n/summationdisplay
i=1ai/parenrightBigg2
=(n−1)2A2.
On the other hand,
/summationdisplay
i/negationslash=jaiaj=/parenleftBiggn/summationdisplay
i=1ai/parenrightBigg2
−/parenleftBiggn/summationdisplay
i=1a2
i/parenrightBigg
=A2−A.
Solution 3.97. Without loss of generality, take a1≤···≤ an.L e tdk=ak+1−ak
fork=1,…,n .T h e n d=d1+···+dn−1.F o r i<j we have that |ai−aj|=
aj−ai=di+···+dj−1. Then,

4.3 Solutions to the problems in Chapter 3 195
s=/summationdisplay
i<j|ai−aj|=n/summationdisplay
j=2j−1/summationdisplay
i=1(di+···+dj−1)
=n/summationdisplay
j=2(d1+2d2+···+(j−1)dj−1)
=(n−1)d1+(n−2)2d 2+···+1·(n−1)dn−1
=n−1/summationdisplay
k=1k(n−k)dk.
Since k(n−k)≥(n−1) (because ( k−1)(n−k−1)≥0) and 4 k(n−k)≤n2
(from the AM-GM inequality), we obtain ( n−1)d≤s≤n2d
4.
In order to see when the equality on the left holds, notice that k(n−k)=
(n−1)⇔n(k−1) =k2−1⇔k=1o r k=n−1, so that ( n−1)d=sonly if
d2=···=dn−2=0 ,t h a ti s , a1≤a2=···=an−1≤an.
For the second equality notice that 4 k(n−k)=n2⇔k=n−k.I fnis
odd, the equality 4 k(n−k)=n2holds only when dk=0f o ra l l k, therefore
a1=···=an=0 .I f nis even, say n=2k, then only dkcan be diﬀerent from
zero and then a1=···=ak≤ak+1=···=a2k.
Solution 3.98. Consider the polynomial P(t)=tb(t2−b2)+bc(b2−c2)+ct(c2−t2).
This satisﬁes the identities P(b)=P(c)=P(−b−c) = 0, therefore P(t)=
(b−c)(t−b)(t−c)(t+b+c), since the coeﬃcient of t3is (b−c). Hence
/vextendsingle/vextendsingleab(a2−b2)+bc(b2−c2)+ca(c2−a2)/vextendsingle/vextendsingle=|P(a)|
=|(b−c)(a−b)(a−c)(a+b+c)|.
The problem is to ﬁnd the least number Msuch that the following inequality holds
for all numbers a,b,c:
|(a−c)(a−b)(b−c)(a+b+c)|≤M(a2+b2+c2)2.
If (a,b,c) satisﬁes the inequality, then ( λa, λb, λc ) also satisﬁes it for any real
number λ. Therefore, we can assume, without loss of generality, that a2+b2+c2=
1. In this way the problem becomes the search for the maximum value of P=
|(a−b)(a−c)(b−c)(a+b+c)|for real numbers a,b,csuch that a2+b2+c2=1 .
Note that
[3(a2+b2+c2)]2=[ 2 (a−b)2+2 (a−c)(b−c)+(a+b+c)2]2
≥8|(a−c)(b−c)|[2(a−b)2+(a+b+c)2]
≥16√
2|(a−c)(b−c)(a−b)(a+b+c)|
=1 6√
2P
The two inequalities are obtained using the AM-GM inequality.

196 Solutions to Exercises and Problems
Thus, P≤9
16√
2, and the maximum value is9
16√
2because the equality holds
witha=3√
3+√
6
6√
2,b=√
6
6√
2andc=√
6−3√
3
6√
2.
Solution 3.99. Fora=2 ,b=c=1
2andn≥3, the inequality is not true.
Ifn= 1, the inequality becomes abc≤1, which follows from3√
abc≤a+b+c
3=1 .
For the case n=2 ,l e t x=ab+bc+ca;n o ws i n c e a2+b2+c2=(a+b+c)2−2(ab+bc+
ca)=9 −2xandx2=(ab+bc+ca)2≥3(a2bc+ab2c+abc2)=3abc(a+b+c)=9abc,
the inequality is equivalent to abc(9−2x)≤3, but it will be enough to prove that
x2(9−2x)≤27. This last inequality is in turn equivalent (2 x+3 ) (x−3)2≥0.
Solution 3.100. First, the AM-GM inequality leads us to ca+c+a≥33√
c2a2.
From this we get
(a+1 ) (b+1 )2
33√
c2a2+1≥(a+1 ) (b+1 )2
ca+c+a+1=(a+1 ) (b+1 )2
(c+1 ) (a+1 )=(b+1 )2
(c+1 ).
Similarly for the other two terms of the sum; therefore
(a+1 ) (b+1 )2
33√
c2a2+1+(b+1 ) (c+1 )2
33√
a2b2+1+(c+1 ) (a+1 )2
33√
b2c2+1
≥(b+1 )2
(c+1 )+(c+1 )2
(a+1 )+(a+1 )2
(b+1 ).
Now, apply inequality (1.11).
Solution 3.101. Using Ravi’s transformation a=x+y,b=y+z,c=z+x,w e
ﬁnd that x+y+z=3
2andxyz≤(x+y+z
3)3=1
8.M o r e o v e r ,
a2+b2+c2+4abc
3=(a2+b2+c2)(a+b+c)+4abc
3
=2((y+z)2+(z+x)2+(x+y)2)(x+y+z)+4 (y+z)(z+x)(x+y)
3
=4
3((x+y+z)3−xyz)
≥4
3/parenleftBigg/parenleftbigg3
2/parenrightbigg3
−1
8/parenrightBigg
=13
3.
Therefore the minimum value is13
3.
Solution 3.102. Apply Ravi’s transformation a=y+z,b=z+x,c=x+y,s o
that the inequality can be rewritten as
(2z)4
(z+x)(2x)+(2x)4
(x+y)(2y)+(2y)4
(y+z)(2z)
≥(y+z)(z+x)+(z+x)(x+y)+(x+y)(y+z).

4.3 Solutions to the problems in Chapter 3 197
From inequality (1.11) and Exercise 1.27, we obtain
(2z)4
(z+x)(2x)+(2x)4
(x+y)(2y)+(2y)4
(y+z)(2z)≥8(x2+y2+z2)2
x2+y2+z2+xy+yz+zx
≥8(x2+y2+z2)2
2(x2+y2+z2).
On the other hand, ( y+z)(z+x)+(z+x)(x+y)+(x+y)(y+z)=3 ( xy+
yz+zx)+(x2+y2+z2); then it is enough to prove that 4( x2+y2+z2)≥
3(xy+yz+zx)+(x2+y2+z2), which can be reduced to x2+y2+z2≥xy+yz+zx.
Solution 3.103. The substitution x=a+b
a−b,y=b+c
b−c,z=c+a
c−ahas the property
thatxy+yz+zx= 1. Using the Cauchy-Schwarz inequality, ( x+y+z)2≥
3(xy+yz+zx) = 3, therefore |x+y+z|≥√
3>1.
Solution 3.104. It will be enough to consider the case x≤y≤z.T h e n x=y−a,
z=y+bwitha,b≥0.
On the one hand, we have xz=1−xy−yz=1−(y−a)y−y(y+b)=
1−2y2+ay−byand on the other, xz=(y−a)(y+b)=y2−ay+by−ab.
Adding both identities, we get 2xz =1−y2−ab,s ot h a t2 xz−1=−y2−ab≤0.
If 2xz=1 ,t h e n y=0a n d xz= 1, a contradiction, therefore xz <1
2.
The numbers x=y=1
nandz=1
2(n−1
n)s a t i s f y x≤y≤zandxy+yz+zx=
1. However, xz=1
2n(n−1
n)=1
2−1
2n2can be as close as we wish to1
2, therefore,
the value1
2cannot be improved.
Solution 3.105. Suppose that a=[x]a n dt h a t r={x}. Then, the inequality is
equivalent to
/parenleftbigga+2r
a−a
a+2r/parenrightbigg
+/parenleftbigg2a+r
r−r
2a+r/parenrightbigg
>9
2.
This inequality reduces to
2/parenleftBigr
a+a
r/parenrightBig
−/parenleftbigga
a+2r+r
2a+r/parenrightbigg
>5
2.
But sincer
a+a
r≥2, it is enough to prove that
a
a+2r+r
2a+r<3
2.
Buta+2r≥a+rand 2a+r≥a+r; moreover, the two equalities cannot hold
at the same time (otherwise a=r= 0), therefore
a
a+2r+r
2a+r<a
a+r+r
a+r=1<3
2.

198 Solutions to Exercises and Problems
Solution 3.106. Inequality (1.11) shows that
a+b+c≥1
a+1
b+1
c≥32
a+b+c,
so thata+b+c
3≥3
a+b+c. Thus, it is enough to prove that a+b+c≥3
abc.
Since ( x+y+z)2≥3(xy+yz+zx), we have
(a+b+c)2≥/parenleftbigg1
a+1
b+1
c/parenrightbigg2
≥3/parenleftbigg1
ab+1
bc+1
ca/parenrightbigg
=3
abc(a+b+c),
and from here it is easy to conclude the proof.
Solution 3.107. By means of the Cauchy-Schwarz inequality we get
(a+b+1 ) (a+b+c2)≥(a+b+c)2.
Then
a+b+c2
(a+b+c)2+a2+b+c
(a+b+c)2+a+b2+c
(a+b+c)2
≥1
a+b+1+1
b+c+1+1
c+a+1≥1.
Therefore,
2(a+b+c)+(a2+b2+c2)≥(a+b+c)2=a2+b2+c2+2 (ab+bc+ca),
and the result follows.Solution 3.108. F o ra ni n t e r i o rp o i n t PofABC, consider the point Qon the
perpendicular bisector of BCsatisfying AQ=AP.L e t Sbe the intersection
ofBPwith the tangent to the circle at Q. Then, SP+PC≥SC, therefore
BP+PC=BS+SP+PC≥BS+SC.
On the other hand, BS+SC≥BQ+QC,t h e n BP+PCis minimum if
P=Q.
LetTbe the midpoint of MN. Since the triangle AMQ is isosceles and MT
is one of its altitudes, then MT=ZQwhere Zis the foot of the altitude of Q
overAB.T h e n MN+BQ+QC=2 (MT+QC)=2 ( ZQ+QC) is minimum
when Z,Q,Care collinear and this means CZis the altitude. By symmetry, BQ
should be also an altitude and then Pis the orthocenter.
Solution 3.109. LetHbe the orthocenter of the triangle MNP ,a n dl e t A
/prime,B/prime,C/prime
be the projections of HonBC,CA,AB, respectively. Since the triangle MNP
is acute, Hbelongs to the interior of the triangle MNP ; hence, it belongs to the
interior of the triangle ABC too, and therefore
x≤HA/prime+HB/prime+HC/prime≤HM+HN+HP≤2X.

4.3 Solutions to the problems in Chapter 3 199
The second inequality is evident, the other two will be presented as the following
two lemmas.
Lemma 1 .I fHis an interior point or belongs to the sides of a triangle ABC,
and if A/prime,B/prime,C/primeare its projections on BC,CA,AB, respectively, then x≤
HA/prime+HB/prime+HC/prime,w h e r e xis the length of the shortest altitude of ABC.
Proof.
HA/prime+HB/prime+HC/prime
x≥HA/prime
ha+HB/prime
hb+HC/prime
hc=(BHC )
(ABC)+(CHA)
(ABC)+(AHB)
(ABC)=1.
/square
Lemma 2 .I fMNP is an acute triangle and His its orthocenter, then HM+HN+
HP≤2X,w h e r e Xis the length of the largest altitude of the triangle MNP .
Proof. Suppose that ∠M≤∠N≤∠P,t h e n NP≤PM≤MNand so it happens
thatXis equal to the altitude MM/prime. We need to prove that HM+HN+HP≤
2MM/prime=2 (HM+HM/prime) or, equivalently, that HN+HP≤HM+2HM/prime./square
LetH/primebe the symmetric point of Hwith respect to NP;s i n c e MNH/primePis
a cyclic quadrilateral, Ptolemy’s theorem tells us that
H/primeM·NP=H/primeN·MP+H/primeP·MN≥H/primeN·NP+H/primeP·NP,
and then we get H/primeN+H/primeP≤H/primeM=HM+2HM/prime.
Solution 3.110. Without loss of generality, we can suppose that x≤y≤z.T h e n
x+y≤z+x≤y+z,xy≤zx≤yz,2z2(x+y)≥2y2(z+x)≥2×2(y+z),
1
√
2z2(x+y)≤1
√
2y2(z+x)≤1
√
2×2(y+z). If we resort to the rearrangement inequality
and apply it twice, we have
/summationdisplay 2yz
/radicalbig
2×2(y+z)≥/summationdisplay xy+zx
/radicalbig
2×2(y+z).
Now, adding/summationtext2x2
√
2×2(y+z)to both sides of the last inequality, we obtain
/summationdisplay2x2+2yz
/radicalbig
2×2(y+z)≥/summationdisplay2x2+xy+zx
/radicalbig
2×2(y+z)
=/summationdisplay2x2+x(y+z)
/radicalbig
2×2(y+z)
≥/summationdisplay2/radicalbig
2×3(y+z)
/radicalbig
2×2(y+z)
=2 (√
x+√
y+√
z)=2.

200 Solutions to Exercises and Problems
Second solution. First, note that
x2+yz
/radicalbig
2×2(y+z)=x2−x(y+z)+yz
/radicalbig
2×2(y+z)+x(y+z)
/radicalbig
2×2(y+z)
=(x−y)(x−z)
/radicalbig
2×2(y+z)+/radicalbigg
y+z
2
≥(x−y)(x−z)
/radicalbig
2×2(y+z)+√
y+√
z
2.
Similarly for the other two elements of the sum; then
/summationdisplay x2+yz
/radicalbig
2×2(y+z)≥/summationdisplay(x−y)(x−z)
/radicalbig
2×2(y+z)+√
x+√
y+√
z.
Then, it is enough to prove that
(x−y)(x−z)
/radicalbig
2×2(y+z)+(y−z)(y−x)
/radicalbig
2y2(z+x)+(z−x)(z−y)
/radicalbig
2z2(x+y)≥0.
Without loss of generality, suppose that x≥y≥z.T h e n(x−y)(x−z)
√
2×2(y+z)≥0, and
(y−z)(y−x)
/radicalbig
2y2(z+x)+(z−x)(z−y)
/radicalbig
2z2(x+y)
=(x−z)(y−z)
/radicalbig
2z2(x+y)−(y−z)(x−y)
/radicalbig
2y2(z+x)≥(x−y)(y−z)
/radicalbig
2z2(x+y)−(y−z)(x−y)
/radicalbig
2y2(z+x)
=(y−z)(x−y)/parenleftBigg
1
/radicalbig
2z2(x+y)−1
/radicalbig
2y2(z+x)/parenrightBigg
≥0.
The last inequality is a consequence of having y2(z+x)=y2z+y2x≥yz2+z2x=
z2(x+y).
Solution 3.111. Inequality (1.11) leads to
a2
2+b+c2+b2
2+c+a2+c2
2+a+b2≥(a+b+c)2
6+a+b+c+a2+b2+c2.
Then, we need to prove that 6+ a+b+c+a2+b2+c2≤12, but since a2+b2+c2=3 ,
it is enough to prove that a+b+c≤3. But we also have ( a+b+c)2=a2+b2+
c2+2 (ab+bc+ca)≤3(a2+b2+c2)=9 .
The equality holds if and only if a=b=c=1 .
Solution 3.112. First, note that
1−a−bc
a+bc=2bc
1−b−c+bc=2bc
(1−b)(1−c)=2bc
(c+a)(a+b).

4.3 Solutions to the problems in Chapter 3 201
Then, the inequality is equivalent to
2bc
(c+a)(a+b)+2ca
(a+b)(b+c)+2ab
(b+c)(c+a)≥3
2.
This last inequality can be simpliﬁed to
4[bc(b+c)+ca(c+a)+ab(a+b)]≥3(a+b)(b+c)(c+a),
which in turn is equivalent to the inequality
ab+bc+ca≥9abc.
But this inequality follows from ( a+b+c)(1
a+1
b+1
c)≥9.
Solution 3.113. Notice that ( x√
y+y√
z+z√
x)2=x2y+y2z+z2x+2 (xy√
yz+
yz√
zx+zx√
xy).
The AM-GM inequality implies that
xy√
yz=√
xyz/radicalbig
xy2≤xyz+xy2
2,
then
(x√
y+y√
z+z√
x)2≤x2y+y2z+z2x+xy2+yz2+zx2+3xyz.
Since ( x+y)(y+z)(z+x)=x2y+y2z+z2x+xy2+yz2+zx2+2xyz,w eo b t a i n
(x√
y+y√
z+z√
x)2≤(x+y)(y+z)(z+x)+xyz
≤(x+y)(y+z)(z+x)+1
8(x+y)(y+z)(z+x)
=9
8(x+y)(y+z)(z+x).
Therefore K2≥9
8,a n dt h e n K≥3
2√
2.W h e n x=y=z, the equality holds with
K=3
2√
2, hence this is the minimum value.
Second solution. Apply the Cauchy-Schwarz inequality in the following way:
x√
y+y√
z+z√
x=√
x√
xy+√
y√
yz+√
z√
zx≤/radicalbig
(x+y+z)(xy+yz+zx).
After that, use the AM-GM inequality several times to produce
(x+y+z)
3(xy+yz+zx)
3≤3√
xyz3/radicalbig
x2y2z2=xyz≤(x+y)
2(y+z)
2(z+x)
2.

202 Solutions to Exercises and Problems
Solution 3.114. The left-hand side of the inequality can be written as
a2b2cd+ab2c2d+abc2d2+a2bcd2+a2bc2d+ab2cd2=abcd(ab+bc+cd+ac+ad+bd).
The AM-GM inequality implies that a2b2c2d2≤(a2+b2+c2+d2
4)4=/parenleftbig1
4/parenrightbig4, hence
abcd≤1
16. To see that the factor ( ab+bc+cd+ac+ad+bd)i sl e s st h a n3
2we
can proceed in two forms.
The ﬁrst way is to apply the Cauchy-Schwarz inequality to obtain
(ab+bc+cd+ac+ad+bd+ba+cb+dc+ca+da+db)
≤(a2+b2+c2+d2+a2+b2+c2+d2+a2+b2+c2+d2)=3.
The second way consists in applying the AM-GM inequality as follows:
(ab+bc+cd+ac+ad+bd)
≤a2+b2
2+b2+c2
2+c2+d2
2+a2+c2
2+a2+d2
2+b2+d2
2=3
2.
Solution 3.115. (a) After some algebraic manipulation and some simpliﬁcations
we obtain
(1 +x+y)2+( 1+ y+z)2+( 1+ z+x)2
=3+4 ( x+y+z)+2 (xy+yz+zx)+2 ( x2+y2+z2).
Now, the AM-GM inequality implies that
(x+y+z)≥33√
xyz≥3,
(xy+yz+zx)≥33/radicalbig
x2y2z2≥3,
(x2+y2+z2)≥33/radicalbig
x2y2z2≥3.
Then, (1 + x+y)2+( 1+ y+z)2+( 1+ z+x)2≥3+4· 3+2· 3+2· 3 = 27.
The equality holds when x=y=z=1 .
(b) Again, after simpliﬁcation, the inequality is equivalent to
3+4 ( x+y+z)+2 (xy+yz+zx)+2 (x2+y2+z2)
≤3(x2+y2+z2)+6 (xy+yz+zx)
and also to 3 + 4 u≤u2+2v,w h e r e u=x+y+z≥3a n d v=xy+yz+zx≥3.
Butu≥3 implies that ( u−2)2≥1, then ( u−2)2+2v≥1+6=7 .
The equality holds when u=3a n d v=3 ,t h a ti s ,w h e n x=y=z=1 .
Solution 3.116. Notice that
1
1+a2(b+c)=1
1+a(ab+ac)=1
1+a(3−bc)=1
3a+1− abc.

4.3 Solutions to the problems in Chapter 3 203
The AM-GM inequality implies that 1 =ab+bc+ca
3≥3√
a2b2c2,t h e n abc≤1. Thus
1
1+a2(b+c)=1
3a+1− abc≤1
3a.
Similarly,1
1+b2(c+a)≤1
3band1
1+c2(a+b)≤1
3c. Therefore,
1
1+a2(b+c)+1
1+b2(c+a)+1
1+c2(a+b)≤1
3a+1
3b+1
3c
=bc+ca+ab
3abc=1
abc.
Solution 3.117. The inequality is equivalent to
(a+b+c)/parenleftbigg1
a+b+1
b+c+1
c+a/parenrightbigg
≥k+(a+b+c)k=(a+b+c+1 )k.
On the other hand, using the condition a+b+c=ab+bc+ca,w eh a v e
1
a+b+1
b+c+1
c+a=a2+b2+c2+3 (ab+bc+ca)
(a+b)(b+c)(c+a)
=a2+b2+c2+2 (ab+bc+ca)+( ab+bc+ca)
(a+b)(b+c)(c+a)
=(a+b+c)(a+b+c+1 )
(a+b+c)2−abc.
Hence
(a+b+c)
(a+b+c+1 )/parenleftbigg1
a+b+1
b+c+1
c+a/parenrightbigg
=(a+b+c)2
(a+b+c)2−abc≥1,
and since the equality holds if and only if abc= 0, we can conclude that k=1i s
the maximum value.
Solution 3.118. Multiplying both sides of the inequality by the factor ( a+b+c),
we get the equivalent inequality
9(a+b+c)(a2+b2+c2)+2 7 abc≥4(a+b+c)3,
which in turn is equivalent to the inequality
5(a3+b3+c3)+3abc≥3(ab(a+b)+ac(a+c)+bc(b+c)).
By the Sch¨ ur inequality with n= 1, Exercise 1.83, it follows that
a3+b3+c3+3abc≥ab(a+b)+bc(b+c)+ca(c+a),
and the Muirhead’s inequality tells us that 2[3, 0,0]≥2[2,1,0], which is equivalent
to
4(a3+b3+c3)≥2(ab(a+b)+ac(a+c)+bc(b+c)).
Adding these last inequalities, we get the result.

204 Solutions to Exercises and Problems
Solution 3.119. Lemma .I fa,b>0, then1
(a−b)2+1
a2+1
b2≥4
ab.
Proof. In order to prove the lemma notice that1
(a−b)2+1
a2+1
b2−4
ab=(a2+b2−3ab)2
a2b2(a−b)2.
/square
Without loss of generality, z=m i n {x, y, z}; now apply the lemma with
a=(x−z)a n d b=(y−z), to obtain
1
(x−y)2+1
(y−z)2+1
(z−x)2≥4
(x−z)(y−z).
Now, it is left to prove that xy+yz+zx≥(x−z)(y−z); but this is equivalent
to 2z(y+x)≥z2, which is evident.
Solution 3.120. In the case of part (i), there are several ways to prove it.
First form. We can prove that
x2
(x−1)2+y2
(y−1)2+z2
(z−1)2−1=(yz+zx+xy−3)2
(x−1)2(y−1)2(z−1)2.
Second form . With the substitution a=x
x−1,b=y
y−1,c=z
z−1, the inequality
is equivalent to a2+b2+c2≥1, and the condition xyz= 1 is equivalent to
abc=(a−1)(b−1)(c−1) or ( ab+bc+ca)+1= a+b+c. With the previous
identities we can obtain
a2+b2+c2=(a+b+c)2−2(ab+bc+ca)
=(a+b+c)2−2(a+b+c−1)
=(a+b+c−1)2+1,
therefore
a2+b2+c2=(a+b+c−1)2+1.
Part (ii) can be proved depending on how we prove part (i). For instance, if
we used the second form , the equality holds when a2+b2+c2=1a n d a+b+c=1 .
(In the ﬁrst form , the equality holds when xyz=1a n d xy+yz+zx=3 ) .F r o m
the equations we can cancel out one variable, for instance c(and since c=1−a−b,
if we ﬁnd that aandbare rational numbers, then cwill be a rational number too),
to obtain a2+b2+ab−a−b= 0, an identity that we can think of as a quadratic
equation in the variable bwith roots b=1−a±√
(1−a)(1+3 a)
2, which will be rational
numbers if (1 −a)a n d( 1+3 a) are squares of rational numbers. If a=k
m,t h e n
m−kandm+3kare squares of integers, for instance, if m=(k−1)2+k,t h e n
m−k=(k−1)2andm+3k=(k+1 )2. Thus, the rational numbers a=k
m,
b=m−k+k2−1
2mandc=1−a−b,w h e n kvaries in the integer numbers, are rational
numbers where the equality holds. There are some exceptions, that is, when k=0 ,
1, since the values a= 0 or 1 are not allowed.

Notation
We use the following standard notation:
N the positive integers (natural numbers)
R the real numbers
R+the positive real numbers
⇔ iﬀ, if and only if
⇒ implies
a∈A the element abelongs to the set A
A⊂BA is a subset of B
|x| the absolute value of the real number x
{x} the fractional part of the real number x
[x] the integer part of the real number x
[a,b] the set of real numbers xsuch that a≤x≤b
(a,b) the set of real numbers xsuch that a<x<b
f:[a,b]→R the function fdeﬁned in [ a,b]w i t hv a l u e si n R
f/prime(x) the derivative of the function f(x)
f/prime/prime(x) the second derivative of the function f(x)
detA the determinant of the matrix A/summationtextn
i=1ai the sum a1+a2+···+an/producttextn
i=1ai the product a1·a2···an/producttext
i/negationslash=jai the product of all a1,a2,…,a nexcept aj
max{a ,b,… } the maximum value between a ,b,…
min{a ,b,… } the minimum value between a ,b,…√
x the square root of the positive real number x
n√
x then-th root of the real number x
expx=exthe exponential function/summationtext
cyclicf(a ,b,… ) represents the sum of the function fevaluated
in all cyclic permutations of the variables a ,b,…

206 Solutions to Exercises and Problems
We use the following notation for the section of Muirhead’s theorem:
/summationdisplay
!F(x1,…,x n) the sum of the n! terms obtained from evaluating Fin
all possible permutations of ( x1,…,x n)
(b)≺(a)( b) is majorized by ( a)
[b]≤[a]1
n!/summationdisplay
!xb1
1xb2
2···xbnn≤1
n!/summationdisplay
!xa1
1xa2
2···xann.
We use the following geometric notation:
A, B, C the vertices of the triangle ABC
a,b,c the lengths of the sides of the triangle ABC
A/prime,B/prime,C/primethe midpoints of the sides BC,CA andAB
∠ABC the angle ABC
∠A the angle in the vertex Aor the measure of the angle A
(ABC) the area of the triangle ABC
(ABCD… ) the area of the polygon ABCD…
ma,mb,mcthe lengths of the medians of the triangle ABC
ha,hb,hc the lengths of the altitudes of the triangle ABC
la,lb,lc the lengths of the internal bisectors of the triangle ABC
s the semiperimeter of the triangle ABC
r the inradius of the triangle ABC, the radius of the incircle
R the circumradius of the triangle ABC, the radius of the
circumcircle
I,O,H,G the incenter, circumcenter, orthocenter and centroid
of the triangle ABC
Ia,Ib,Ic the centers of the excircles of the triangle ABC.
We use the following notation for reference of problems:
IMO International Mathematical Olympiad
APMO Asian Paciﬁc Mathematical Olympiad
(country, year) problem corresponding to the mathematical olympiad
celebrated in that country, in that year, in some stage.

Bibliography
[1] Altshiller, N., College Geometry: An Introduction to Modern Geometry of the
Triangle and the Circle. Barnes and Noble, 1962.
[2] Andreescu, T., Feng, Z., Problems and Solutions from Around the World.
Mathematical Olympiads 1999-2000. MAA, 2002.
[3] Andreescu, T., Feng, Z., Lee, G., Problems and Solutions from Around the
World . Mathematical Olympiads 2000-2001. MAA, 2003.
[4] Andreescu, T., Enescu, B., Mathematical Olympiad Treasures. Birkh¨auser,
2004.
[5] Barbeau, E.J., Shawyer, B.L.R., Inequalities. A Taste of Mathematics, vol.
4, 2000.
[6] Bulajich, R., G´ omez Ortega, J.A., Geometr´ ıa. Cuadernos de Olimpiadas de
Matem´ aticas, Instituto de Matem´ aticas, UNAM, 2002.
[7] Bulajich, R., G´ omez Ortega, J.A., Geometr´ ıa. Ejercicios y Problemas .
Cuadernos de Olimpiadas de Matem´ aticas, Instituto de Matem´ aticas, UNAM,
2002.
[8] Courant, R., Robbins, H., ¿Qu´e son las Matem´ aticas? Fondo de Cultura
Econ´omica, 2002.
[9] Coxeter, H., Greitzer, S., Geometry Revisited. New Math. Library, MAA,
1967.
[10] Dorrie, H., 100 Great Problems of Elementary Mathematics. Dover, 1965.
[11] Engel, A., Problem-Solving Strategies. Springer-Verlag, 1998.
[12] Fomin, D., Genkin, S., Itenberg, I., Mathematical Circles. Mathematical
World, Vol. 7. American Mathematical Society, 1996.
[13] Hardy, G.H., Littlewood, J.E., P` olya, G., Inequalities. Cambridge at the
University Press, 1967.
[14] Honsberger, R., Episodes in Nineteenth and Twentieth Century Euclidean
Geometry. New Math. Library, MAA, 1995.

208 Bibliography
[15] Kazarinoﬀ, N., Geometric Inequalities. New Math. Library, MAA. 1961.
[16] Larson, L., Problem-Solving Through Problems. Springer-Verlag, 1990.
[17] Mitrinovic, D., Elementary Inequalities. Noordhoﬀ Ltd., Groningen, 1964.
[18] Niven, I., Maxima and Minima Without Calculus. The Dolciani Math.
Expositions, MAA, 1981.
[19] Shariguin, I., Problemas de Geometr´ ıa.Editorial Mir, 1989.
[20] Soulami, T., Les Olympiades de Math´ ematiques. Ellipses, 1999.
[21] Spivak, M., Calculus. Editorial Benjamin, 1967.

Index
Absolute value, 2
Concavity
Geometric interpretation, 25
Convexity
Geometric interpretation, 25
discrepancy, 46Erd˝os-Mordell
theorem, 81–84, 88
Euler
theorem, 66
Fermat
point, 90, 92
Function
concave, 23
convex, 20quadratic, 4
Greater than, 1
Inequality
arithmetic mean–geometric
m e a n ,9 ,4 7
weighted, 27
Bernoulli, 31
Cauchy-Schwarz, 15, 35
Engel form, 35
Euler, 67H¨older, 27
generalized, 32
harmonic mean–geometric
mean, 8
helpful, 34Jensen, 21
Leibniz, 69
Minkowski, 28Nesbitt, 16, 37, 65
Popoviciu, 32
power mean, 32
Ptolemy, 53
quadratic mean–arithmetic
mean, 19, 36
rearrangement, 13Schur, 31
Tchebyshev, 18
triangle, 3
general form, 3
Young, 27
Leibniz
theorem, 68
Mean
arithmetic, 7, 9, 19, 31geometric, 7, 9, 19, 31h a r m o n i c ,8 ,1 9
power, 32
quadratic, 19
Muirhead
theorem, 43, 44
Ortic
triangle, 95, 98
Pappus
theorem, 80
Pedal
triangle, 99

210 Index
Problem
Fagnano, 88, 94
Fermat-Steiner, 88Heron, 92
with a circle, 93
Pompeiu, 53
Real line, 1
Smaller than, 1
Smaller than or equal to, 2
Solution
Fagnano problem
Fej´er L., 96
Schwarz H., 96
Fermat-Steiner problem
Hofmann-Gallai, 91
Steiner, 92, 94Torricelli, 88, 90
Transformation
Ravi, 55, 73
Viviani
lemma, 88, 90

Copyright Notice

© Licențiada.org respectă drepturile de proprietate intelectuală și așteaptă ca toți utilizatorii să facă același lucru. Dacă consideri că un conținut de pe site încalcă drepturile tale de autor, te rugăm să trimiți o notificare DMCA.

Acest articol: A Mathematical Olympiad ApproachRadmila Bulajich Manfrino [604432] (ID: 604432)

Dacă considerați că acest conținut vă încalcă drepturile de autor, vă rugăm să depuneți o cerere pe pagina noastră Copyright Takedown.

A Mathematical Olympiad ApproachRadmila Bulajich Manfrino [604432]

Copyright Notice

ARGUMENT………………………………………………………………………………………………….. 5 PARTEA I CAP. I – SCURTĂ… [304014]

FORMA DE INVATAMANT CU FRECVENTA LUCRARE DE LICENTA COORDONATOR STIINTIFIC LECT. UNIV. DR. DINU SASU ABSOLVENT DINU CRISTIAN ORADEA 2018… [613274]

SPECIALIZAREA PEDAGOGIA ÎNVĂȚĂMÂNTULUI PRIMAR ȘI PREȘCOLAR [603428]

Străbunicului meu, Iftimie , [622972]

ϹADRUL oΤΕΟ RΕΤΙϹ oΙΝΤRΟDUϹΤΙV AL LU ϹRĂR ΙΙoo [613076]

Să se scrie un program Java care să traseze grafic conica de ecuație generală [620822]

Copyright Notice

Similar Posts