Straight from the boook

The topics chosen reflect those from the first twelve chapters of the previous book: so we have Cauchy-Schwarz, Algebraic Number Theory, Formal Series, Lagrange Interpolation, to name bu

Straight from the Book

Titu Andreescu Gabriel Dospinescu

University of Texas at Dallas Ecole Normale Superieure, Lyon

Library of Congress Control Number: 2012951362

ISBN-13: 978-0-9799269-3-8 ISBN-IO: 0-9799269-3-9

© 2012 XYZ Press, LLC

All rights reserved This work may not be translated or copied in whole or in

part without the written permission of the publisher (XYZ Press, LLC, 3425

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to proprietary rights

9 8 7 6 5 4 3 2 1

www.awesomemath.org

Cover design by Iury Ulzutuev

The only way to learn mathematics is to do mathematics ,

-Paul Halmos

Foreword

This book is a follow-on from the authors' earlier book, 'Problems from the Book' However, it can certainly be read as a stand-alone book: it is not vital to have read the earlier book

The previous book was based around a collection of problems In contrast, this book is based around a collection of solutions These are solutions to some of the (often extremely challenging) problems from the earlier book The topics chosen reflect those from the first twelve chapters of the previous book: so we have Cauchy-Schwarz, Algebraic Number Theory, Formal Series, Lagrange Interpolation, to name but a few

The book is one of the most remarkable mathematical texts I have ever seen First of all, there is the richness of the problems, and the huge variety

of solutions The authors try to give several solutions to each problem, and moreover give insight about why each proof is the way it is, in what way the solutions differ from each other, and so on The amount of work that has been put in, to compile and interrelate these solutions, is simply staggering There

is enough here to keep any devotee of problems going for years and years Secondly, the book is far more than a collection of solutions The solu-tions are used as motivation for the introduction of some very clear expositions

of mathematics And this is modern, current, up-to-the-minute mathematics For example, a discussion of Extremal Graph Theory leads to the celebrated Szemen§di-Trotter theorem on crossing numbers, and to the amazing applica-tions of this by Szekely and then on to the very recent sum-product estimates

of Elekes, Bourgain, Katz, Tao and others This is absolutely state-of-the-art material It is presented very clearly: in fact, it is probably the best exposition

of this that I have seen in print

viii

As another example, the Cauchy-Schwarz section leads on to the

devel-opements of sieve theory, like the Large Sieve of Linnik and the Tunin-Kubilius

inequality And again, everything is incredibly clearly presented The same

applies to the very large sections on Algebraic Number Theory, on p-adic

Analysis, and many others

It is quite remarkable that the authors even know so much current

math-ematics I do not think any of my colleagues would be so well-informed over

so wide an area It is also remarkable that, at least in the areas in which I

am competent to judge, their explanations of these topics are polished and

exceptionally well thought-out: they give just the right words to help someone

understand what is going on

Overall, this seems to me like an 'instant classic' There is so much

material, of such a high quality, wherever one turns Indeed, if one opens the

book at random (as I have done several times), one is pulled in immediately

by the lovely exposition Everyone who loves mathematics and mathematical

thinking should acquire this book

lack of space In addition, the statements of the proposed problems contained' typos and some elementary mistakes which needed further editing Finally, the problems were also considered to be quite difficult to tackle With these points in mind, we came up with a two-part plan: to correct the identified errors and to publish comprehensive solutions to the problems

The first task, editing the statements of the proposed problems, was ple and has already been completed in the second edition of Problems from the Book Although we focused on changing several problems, we also introduced

sim-many new ones The second task, providing full solutions, however, proved to

be more challenging than expected, so we asked for help We created a forum

on www.mathlinks.ro (a familiar site for problem-solving enthusiasts) where solutions to the proposed problems were gathered It was a great pleasure to witness the passion with which some of the best problem-solvers on mathlinks

attacked these tough-nuts This new book is the result of their common effort, and we thank them

Providing solutions to every problem within the limited space of one ume turned out to be an optimistic plan Only the solutions to problems from the first 12 chapters of the second edition of Problems from the Book are

vol-presented here Furthermore, many of the problems are difficult and require

a rather extensive mathematical background We decided, therefore, to plement the problems and solutions with a series of addenda, using various problems as starting points for excursions into "real mathematics" Although

com-x

we never underestimated the role of problem solving, we strongly believe that

the reader will benefit more from embarking on a mathematical journey rather

than navigating a huge list of scattered problems This book tries to reconcile

problem-solving with "professional mathematics"

Let us now delve into the structure of the book which consists of 12

chap-ters and presents the problems and solutions proposed in the corresponding

chapters of Problems from the Book, second edition Many of these problems

are fairly difficult and different approaches are presented for a majority of

them At the end of each chapter, we acknowledge those who provided

so-lutions Some chapters are followed by one or two addenda, which present

topics of more advanced mathematics stemming from the elementary topics

discussed in the problems of that chapter

The first two chapters focus on elementary algebraic inequalities (a

no-table caveat is that some of the problems are quite challenging) and there is

not much to comment regarding these chapters except for the fact that

al-gebraic inequalities have proven fairly popular at mathematics competitions

To provide relief from this rather dry landscape (the reader will notice that

most of the problems in these two chapters start with "Let a, b, c be positive

real numbers"), we included an addendum presenting deep applications of the

Cauchy-Schwarz inequality in analytic number theory For instance, we

dis-cuss Gallagher's sieve, Linnik's large sieve and its version due to Montgomery

We apply these results to the distribution of prime numbers (for instance

Brun's famous theorem stating that the sum of the inverses of the twin primes

converges) The note-worthy Tunin-Kubilius inequality and its classical

appli-cations (the Hardy-Ramanujan theorem on the distribution of prime factors of

n, Erdos' multiplication table problem, Wirsing's generalization of the prime

number theorem) are also discussed The reader will be exposed to the power

of the Cauchy-Schwarz inequality in real mathematics and, hopefully, will

un-derstand that the gymnastics of three-variables algebraic inequalities is not

the Holy Grail

Chapter 3 discusses problems related to the unique factorization of

inte-gers and the p-adic valuation maps Among the topics discussed, we cover

the local-global principle (which is extremely helpful in proving divisibilities

or arithmetic identities), Legendre's formula giving the p-adic valuation of n!,

xi

a beneficial elementary result called lifting the exponent lemma, as well as

more advanced techniques from p-adic analysis One of the most beautiful sults presented in this chapter in the celebrated Skolem-Mahler-Lech theorem concerning the zeros of a linearly recursive sequence Readers will perhaps appreciate the applications of p-adic analysis, which is covered extensively in

re-a long re-addendum This re-addendum discusses, from re-a foundre-ationre-al level, the arithmetic of p-adic numbers, a subject that plays a central role in modern number theory Once the basic groundwork has been laid, we discuss the p-adic analogues of classical functions (exponential, logarithm, Gamma func-tion) and their applications to difficult congruences (for instance, Kazandzidis' famous supercongruence) This serves as a good opportunity to explore the arithmetic of Bernoulli numbers, Volkenborn's theory of p-adic integration, Mahler and Amice's classical theorems characterizing continuous and locally analytic functions on the ring of p-adic integers or Morita's construction of the p-adic Gamma function A second addendum to this chapter discusses various classical estimates on prime numbers, which are used throughout the book Chapter 4 discusses problems and elementary topics related to prime num-

bers of the form 4k + 1 and 4k + 3 The most intriguing problem discussed

is, without any doubt, Cohn's renowned theorem characterizing the perfect squares in the Lucas sequence Chapter 5 is dedicated again to the yoga of algebraic inequalities and is followed by an addendum discussing applications

of Holder's inequality

Chapter 6 focuses on extremal graph theory Most of the problems revolve around Turan's theorem, however the reader will also be exposed to topics such as chromatic numbers, bipartite graphs, etc This chapter is followed

by a relatively advanced addendum, discussing various topics related to the Szemeredi-Trotter theorem, which gives bounds for the number of incidences between a set of points and a set of curves We discuss the theorem's classical probabilistic proof, its generalization to multi-graphs due to Szekely and its application to the sum-product problem due to Elekes, as well as more recent developments due to Bourgain, Katz, and Tao These results are then applied

to natural and nontrivial geometric questions (for instance: what is the least number of distinct distances determined by n points in the plane? what is the maximal number of triangles of the same area?) Finally, another addendum

Xll

completes this chapter and is dedicated to the powerful probabilistic method

After a short discussion on finite probability spaces, we provide many examples

of combinatorial applications

Chapter 7 involves combinatorial and number theoretic applications of

finite Fourier analysis The central principles of this chapter include the roots

of unity and the fact that congruences between integers can be expressed in

terms of sums of powers of roots of unity To provide the reader with a broader

view, we beefed-up this chapter with an addendum discussing Fourier analysis

on finite abelian groups and applying it to Gauss sums of Dirichlet characters,

additive problems, combinatorial or analytic number theory For instance,

the reader will find a discussion of the P6lya-Vinogradov inequality and

Vina-gradov's beautiful use of this inequality to deduce a rather strong bound on

the least quadratic non-residue mod p At the same time, we explore

Dirich-let's L-functions, culminating in a proof of DirichDirich-let's theorem on arithmetic

progressions This section is structured to first present the usual analytic proof

(since it is really a masterpiece from all points of view), up to some important

simplification due to Monsky At the same time, we also discuss how to turn

this into an elementary proof that avoids complex analysis and dramatically

uses Abel's summation formula

Chapter 8 focuses on diverse applications of generating functions This

is an absolutely crucial tool in combinatorics, be it additive or enumerative

In this section, the reader will have the opportunity to explore enumerative

problems (Catalan's problem, counting the number of solutions of linear

dio-phantine equations or the number of irreducible polynomials mod p, of fixed

degree) We also discuss exotic combinatorial identities or recursive sequences,

which can be solved elegantly using generating functions, but also rather

chal-lenging congruences that appear so often in number theory The chapter is

followed by an addendum presenting a very classical topic in enumerative

combinatorics, Lagrange's inversion formula Among the applications, let us

mention Abel's identity, other derived combinatorial identities, Cayley's

the-orem on labeled trees, and various related problems

Chapter 9 is rather extensive, due to the vast nature of the topic covered,

algebraic number theory While we are able only to scratch the surface,

nev-ertheless the reader will find a variety of intriguing techniques and ideas For

xiii

instance, we discuss arithmetic properties of cyclotomic polynomials ing Mann's beautiful theorem on linear equations in roots of unity), rationality problems, and various applications of the theorem of symmetric polynomials

(includ-In addition, we present techniques rooted in the theory of ideals in number fields, finite fields and p-adic methods We also give an overview of the el-ementary algebraic number theory in the addendum following this chapter After a brief review on ideals, field extensions, and algebraic numbers, we proceed with a discussion of the primitive element theorem and embeddings

of number fields into C We also briefly survey Galois theory, and the damental theorem on the prime factorization of ideals in number fields, due

fun-to Kummer and Dedekind Once the foundation has been set, we discuss the prime factorization in quadratic and cyclotomic fields and apply these tech-niques to basic problems that explore the aforementioned theories Finally, we discuss various applications of Bauer's theorem and of Chebotarev's theorem The next addendum is concerned with the fascinating topic of counting the

number of solutions modulo p of systems of polynomial equations We use

this as an opportunity to state and prove the basic structural results on finite fields and introduce the Gauss and Jacobi sums We go ahead and count the number of points over a finite field of a diagonal hypersurface and to compute its zeta function This is a beautiful theorem of Weil, the very tip of a massive iceberg

Chapter 10 focuses on the arithmetic of polynomials with integer cients An essential aspect of the discussion concentrates on Mahler expan-sions, the theory of finite differences, and their applications The techniques used in this chapter are rather diverse Although the problems can be consid-ered basic, they are challenging and require advanced problem-solving skills Chapter 11 provides respite from the difficult tasks mentioned above It

coeffi-discusses Lagrange's interpolation formula, allowing a unified presentation of various estimates on polynomials

The longest and certainly most challenging chapter is the last one It

explores several algebraic techniques in combinatorics The methods are dard but powerful The last part of the chapter deals with applications to geometry, presenting some of Dehn's wonderful ideas The last problem pre-sented in the book is the famous Freiling, Laczkovich, Rinne, Szekeres theorem,

stan-xiv

a stunningly beautiful application of algebraic combinatorics

We would like to thank, again, the members of the mathlinks site for

their invaluable contribution in providing solutions to many of the problems

in this book Special thanks are due to Richard Stong, who did a remarkable

job by pointing out many inaccuracies and suggesting numerous alternative

solutions We would also like to thank Joshua Nichols-Barrer, Kathy Cordeiro

and Radu Sorici, who gave the manuscript a readable form and corrected

several infelicities Many of the problems and results in this book were used by

the authors in courses at the AwesomeMath Summer Program, and students'

reactions guided us in the process of simplifying or adding more details to the

discussed problems We wish to thank them all, for their courage in taking

and sticking with these courses, as well as for their valuable suggestions

Titu Andreescu

titu.andreescu@utdallas.edu

Gabriel Dospinescu gdospi2002@yahoo.com

Contents

1 Some Useful Substitutions 1.1 The relation a 2 + b 2 + e2 = abc + 4 1.2 The relations abc = a + b + e + 2 and ab + be + ea + 2abe = 1 1.3 The relation a 2 + b 2 + e2 + 2abe = 1

3.4 Problems with combinatorial and valuation-theoretic aspects 104

xvi

5 T 2 's Lemma

5.1 Notes

Addendum 5.A Holder's Inequality in Action

6 Some Classical Problems in Extremal Graph Theory

6.1 Notes

Addendum 6.A Some Pearls of Extremal Graph Theory

Addendum 6.B Probabilities in Combinatorics

Addendum 7.A Finite Fourier Analysis

8 Formal Series Revisited

Addendum 8.A Lagrange's Inversion Theorem

9 A Little Introduction to Algebraic Number Theory

9.1 Tools from linear algebra

9.2 Cyclotomy

9.3 The gcd trick

9.4 The theorem of symmetric polynomials

9.5 Ideal theory and local methods

9.6 Miscellaneous problems

9.7 Notes

Addendum 9.A Equations over Finite Fields

Addendum 9.B A Glimpse of Algebraic Number Theory

Contents

205

226 227

10 Arithmetic Properties of Polynomials

10.1 The a - blf(a) - f(b) trick 10.2 Derivatives and p-adic Taylor expansions 10.3 Hilbert polynomials and Mahler expansions 10.4 p-adic estimates

10.5 Miscellaneous problems 10.6 Notes

11 Lagrange Interpolation Formula

11.1 Notes

12 Higher Algebra in Combinatorics

12.1 The determinant trick 12.2 Matrices over lF2 12.3 Applications of bilinear algebra 12.4 Matrix equations 12.5 The linear independence trick 12.6 Applications to geometry 12.7 Notes

Chapter 1

Some Useful Substitutions

Let us first recall the classical substitutions that will be used in the lowing problems All of these are discussed in detail in [3], chapter 1 and the reader is invited to take a closer look there

fol-Consider three positive real numbers a, b, c If abc = 1, a classical tution is

1 1 1

a=x+-, b=y+-, c=z+-

2 Chapter 1 Some Useful Substitutions

In the second case one can find an acute-angled triangle ABC such that

a = cos(A), b = cos(B), e = cos(C)

Of course, in practice one often needs to use a mixture of these substitutions

and to be rather familiar with classical identities and inequalities But

expe-rience comes with practice, so let us delve into some exercises and problems

to see how things really work

1.1 The relation a2 + b2 + c2 = abc + 4

We start with an easy exercise, based on the resolution of a quadratic

equation

1 Prove that if a, b, e 2 0 satisfy la2 + b2 + c2 - 41 = abc, then

(a - 2)(b - 2) + (b - 2)(c - 2) + (c - 2)(a - 2) 2 o

Titu Andreescu, Gazeta Matematidi

Proof If max(a, b, c) < 2, then everything is clear, so assume that

Proof The most natural idea is to consider the hypothesis as a quadratic

equation in a, for instance It becomes a2 ± abc + b2 + c2 - 4 = 0, and solving

the equation yields

=Fbe ± J(b2 - 4)(c2 - 4)

1.1 The relation a2 + b2 + e2 = abc + 4

Thus (b2 - 4)(e2 - 4) = (be ± 2a)2, which can also be written as

(b - 2)(e - 2) = (be ± 2a)2 > o

(b+2)(c+2)

-3

Writing similar expressions for the other two variables, we are done 0 The following exercise is trickier and one needs some algebraic skills in order to solve it We present two solutions, neither of which is really easy

2 Find all triples x, y, z of positive real numbers such that

Since abc = 1, we have

so the second equation can be written

The left-hand side is also equal to

1

z=c+

-c

4 Chapter 1 Some Useful Substitutions

because

~ ~ _ a 2 + b 2 _ (2 b2)

b + a - ab - e a +

We deduce that CL a-I) (2::: ab - 1) = 4 Since 2::: a 2': 3 and 2::: ab 2': 3 (by

the AM-GM inequality and the fact that abc = 1), this can only happen if

a = b = e = 1 and thus when x = y = Z = 2 0

Proof If x + y = 2, the second equation yields xy = 4, so that (x - y)2 = -12

which is a contradiction Thus x + y =f 2 and similarly y + z =f 2, z + x =f 2

The second equation yields

Unless x = y = 2, this implies the inequality 2 > x + y If two of the numbers

x, y, z are equal to 2, then trivially so is the third one If not, the previous

argument shows that 2 > x + y, 2> y + z and 2 > x + z But then the second

equation yields

x+y+z> L X ( Y ; z ) =xy+yz+zx=2(x+y+z),

eye

a contradiction Thus, the only solution is x = Y = z = 2 o

The following problem hides under a clever algebraic manipulation a very

simple AM-GM argument The inequality is quite strong, as the reader can

easily see by trying a brute-force approach

3 Prove that if a, b, e 2': 2 satisfy a2 + b2 + e2 = abc + 4, then

a + b + e + ab + ae + be 2': 2J(a + b + e + 3)(a2 + b2 + e2 - 3)

Marian Tetiva

Proof The hypothesis yields the existence of positive real numbers x, y, Z such that

6 Chapter 1 Some Useful Substitutions

Proof The system has solutions if and only if

A simple case analysis shows that the only solutions are k = l = 1000, m = 2

and its permutations The result follows 0

5 Solve in positive integers the equation

(x+2)(y+2)(z+2) = (x+y+z+2)2

Titu Andreescu

Proof A simple algebraic manipulation shows that the equation is equivalent

to x2+y2+z2 = xyz+4 and, seeing this as a quadratic equation in z, we obtain

the equivalent form (x2-4)(y2-4) = (xy_2z)2 If x2 < 4, then y2 < 4 as well

and so x = y = 1, yielding z = 2 If x = 2, then y = z In all other cases, we

can find a positive square-free integer D (which is easily seen to be different

from 1) and positive integers u, v such that x2 - 4 = Du 2 and y2 - 4 = Dv 2

Thus, solving the problem comes down to solving the generalized Pell equation

a 2 - Db 2 = 4, which is a classical topic: this equation always has nontrivial

integer solutions and if (ao, bo) is the smallest solution with ao, bo > 0, then

all solutions are given by

~ ~ Jv [(no + ;vn) n _ ( aD - ;vn) nJ o

Part of the following problem can be dealt in a classical way, but we do not know how to solve it entirely without using the trick of substitutions

6 The sequence (an )n2:0 is defined by ao = al = 97 and

an+l = anan-l + J(a~ - l)(a~_l - 1)

for all n 2: 1 Prove that 2 + y2 + 2a n is a perfect square for all n

Proof Writing

(an+! - anan_d 2 = (a; - l)(a;_l - 1)

and simplifying this expression yields

an-l + an + an+l - anan-lan+l = ,

thus

Titu Andreescu

(2an_l)2 + (2an )2 + (2an+d 2 - (2an-l)(2an )(2an+d = 4

Since we clearly have an > 2 for all n, this implies the existence of a sequence

x n > 1 such that 2an = Xn + x;:;-l and such that Xn+l = XnXn-l Thus logxn

satisfies a Fibonacci-type recursive relation and so we can immediately find out the general term of the sequence (an)n Namely, a small computation shows that if we define a = 2 + yI3, then Xn = a 4Fn , where Fn is the nth Fibonacci number Thus

S Chapter 1 Some Useful Substitutions

Remark 1.1 Here is an alternative proof of the fact that all terms of the

sequence are integers, without the use of substitutions The method that we

will use for this problem appears in many other problems As we saw in the

previous solution, the sequence satisfies the recursive relation

Writing the same relation for n + 1 instead of n and subtracting the two yields

the identity

a~+2 - a~-I = 2anan+l(an+2 - an-I)

Note that (an)n is an increasing sequence (this follows trivially by induction

from the recursive relation), so that we can divide by an+2 - an-I i- 0 in the

previous relation and get an+2 = 2anan+l - an-I The last relation clearly

implies that all terms of the sequence are integers (since one can immediately

check that this is the case with the first three terms of the sequence) Note

however that it does not seem to follow easily that 2 + V2 + 2a n is a perfect

square using this method

Remark 1.2 There are a lot of examples of very complicated recurrence

relations that rather unexpectedly yield integers For instance, the reader

can try to prove the following result concerning Somos-5 sequences: let

al = a2 = = a5 = 1 and let

for n 2: O Then an is an integer for all n Similarly one defines Somos-6,

Somos- 7, etc sequences by the formulas

ao = al = = a5 = 1,

ao = al = = a6 = 1,

etc One can prove (though this is not easy) that all terms of Somos-6 and

Somos- 7 sequences are integers Surprisingly, this fails for Somos-S sequences

(in which case al7 is no longer an integer!)

1.2 The relations abc = a + b + e + 2 and'ab + be + ea + 2abe = 1

9

The first inequality in the following problem is very useful in practice and

we will meet it very often in the following problems

7 Prove that if x, y, z > 0 and xyz = x + y + z + 2, then

a(a - b)(a - c) + b(b - a)(b - c) + e(e - a)(e - b) 2: 0, while the second one follows by adding up the inequalities

10 Chapter 1 Some Useful Substitutions

Proof We will argue by contradiction, assuming that xyz > x + y + z + 2 We

claim that we can find 0 < r < 1 such that X = rx, Y = ry, Z = r z satisfy

XY Z = X + Y + Z + 2 Indeed, this comes down to the vanishing of

f (r) = r3 xy z - r (x + y + z) - 2 between 0 and 1, and this is clear, since f(O) < 0 and f(1) > O Next, the

condition xy + yz + zx = 2(x + y + z) yields

XY + YZ + ZX = 2r(X + Y + Z) < 2(X + Y + Z)

This contradicts the first inequality of problem 7 o

Proof The condition can also be rewritten in the form

(x - 1)(y - 1) + (y - 1)(z - 1) + (z - 1)(x - 1) = 3

or in the form

xyz - x - y - z - 1 = (x - 1)(y - 1)(z - 1)

We will discuss several cases If x, y, z ::::: 1, then by the A~\1-GM inequality

and the first identity, we get

which yields, thanks to the second identity, the desired estimate

If x, y, z ::; 1 or if only one of the numbers x, y, z is smaller than or equal

to 1, then (x - 1)(y - 1)(z - 1) ::; 0 and so xyz ::; x + y + z + 1 in this case

Finally, if two of the numbers are smaller than 1, say x, y ::; 1, the desired

inequality can be written in the form 0 ::; x + y + z(1 - xy) + 2, which is

Proof For three positive real numbers x, y, z consider fixing the first two

el-ementary symmetric polynomials 171 = x + y + z and 172 = xy + yz + zx and

letting 173 = xyz vary This amounts to varying only the constant term in the

polynomial

1.2 The relations abc = a + b + e + 2 and,ab + be + ea + 2abe = 1 11

and defining X,y,z to be the three roots of this polynomial (in some order)

Increasing 173, i.e lowering the constant term, corresponds geometrically to lowering the graph As we lower the graph, the smallest root increases, thus

we maintain three positive real roots until the smallest root becomes a double root If the double root is at t = a and the larger root at t = b, then we have

for 1 < a::; 2 But this rearranges to (a-2)2(a 2 -1) ::::: 0 and we are done 0

The technique used in the first solution of the next problem is rather satile and the reader is invited to read the addendum 5.A for more examples

ver-9 Let x, y, z > 0 be such that xy + yz + zx + xyz = 4 Prove that

Proof Using the usual substitution

12 Chapter 1 Some Useful Substitutions

This is a quite strong inequality and it is easy to convince oneself that

most applications of classical techniques fail However, the following smart

application of Holder's inequality does the job:

so it is enough to prove that

This reduces after expanding to 2: a(b - e)2 2': 0, which is clear 0

Proof First, we get rid of those nasty square roots, via the substitution

and replacing these values in the inequality yields the equivalent form

3(a + b + e)2 2': 16(a + be)(b + ea)(e + ab)

The hypothesis becomes a2 + b2 + e2 + 2abe = 1, so that there exists an

acute-angled triangle ABC such that a = cosA, b = cosB, e = cosC Next,

observe that

e + ab = cosC + cos A cosB = - cos(A + B) + cos A cosB = sin A sinB

Using this (and similar identities obtained by permuting the variables), the

desired inequality becomes

v'3 L cos A 2': 4 II sin A

Using the well-known identities

s = 4R II cos ~ , r = 4R II sin ~ , L cos A = 1 + ~,

1.2 The relations abc = a + b + e + 2 and'ab + be + ea + 2abe = 1

the inequality becomes

v'3r+R> 2rs

13

or (R+r)Rv'3 2': 2rs This splits trivially into R+r 2': 3r (Euler's inequality)

and s ::; 3r R, the last one being well-known and easy 0 Here is yet another easy application of problem 7

10 Let u, v, w > 0 be positive real numbers such that

Then u = be, v = ea and w = ab, so that the inequality becomes a + b + e 2':

ab + be + ea and the hypothesis is ab + be + ea + abc = 4 Another substitution

a =~, x b -- ~ e = ~

yields xyz = x+y+z+2 We need to prove that xy+yz+zx 2': 2(x+y+z),

which is the first component of problem 7 0 The following problem has some common points with problem 7 and one can actually deduce it from that result But the proof is not formal

11 Prove that if a, b, e > 0 and x = a + -b 1 , y = b + ~, z = e + ~, then

Vasile Cartoaje

14 Chapter 1 Some Useful Substitutions

Proof The method is the same as the one used in problem 8 The starting

point is the observation that xyz 2: 2 + x + y + z Indeed,

xyz = abc + - + a + b + e + - + -b + - 2: 2 + x + y + z

Let us assume for sake of contradiction that xy + yz + zx < 2(x + y + z) and

let us choose r E (0,1] such that X = rx, Y = ry, Z = rz satisfy XY Z =

2 + X + Y + Z This equality is equivalent to r3 xy z = 2 + r (x + y + z) and such

r exists by continuity of the function f(r) = r 3 xyz-2-r(x+y+z) and by the

fact that f(I) 2: 0 and f(O) < O The hypothesis xy + yz + zx < 2(x + y + z)

can also be written as

XY + YZ + ZX < 2r(X + Y + Z) ~ 2(X + Y + Z)

This contradicts the first inequality in problem 7 o

Proof As in the previous proof, we obtain that xyz 2: x + y + z + 2 Since we

obviously have min(xy, yz, zx) > 1, this implies that z 2: 2;:yX_+l and similar

inequalities obtained by permuting the variables Next, we have

x + y + z = a + - + b + - + e + - > 6

-In particular, there are two numbers, say x, y, such that x + y 2: 4 The

inequality to be proved is equivalent to z 2: 2(~~~~2XY, so we are done if we

can prove that

2 + x + Y 2(x + y) - xy

But this is equivalent, after an easy computation (in which it is convenient

to denote S = x + y and P = XV), to (xy - y - x)2 2: (x - 2)(y - 2) If

(x - 2)(y - 2) ~ 0, this is clear; otherwise x, y 2: 2 as x + y 2: 4 Let

u = x - 2,v = y - 2, then the inequality becomes (uv + u + v? 2: uv, with

The form of the following inequality strongly suggests the use of the

Cauchy-Schwarz inequality It turns out that this approach works, but in

a rather indirect and mysterious way, which makes the problem rather hard

1.2 The relations abc = a + b + e + 2 and tlb + be + ea + 2abe = 1 15

12 Prove that for all a, b, e > 0,

xy z = x + y + z + 2 we have

-' , -" + + > -

x 2 + 1 y2 + 1 z2 + 1 - 5 Applying the Cauchy-Schwarz inequality (or what is also called Titu's lemma),

we obtain the bound

(x 1)2 (Y-I? (z 1)2 (x+y+z 3)2

x 2 + 1 + y2 + 1 + z2 + 1 2: x 2 + y2 + z2 + 3

It remains to prove that the last quantity is at least ~ Let S = x + y + z and

P = xy + yz + zx, so that the inequality becomes

Notice that as % + ~ 2: 2 and cyclic permutations of this, we have S 2: 6 Also,

by problem 7 we have P 2: 2S Therefore,

(S - 3)2 > (S 3)2 = S - 3 = 1 _ _ 2_ > ~

Remark 1.3 There are other solutions for this problem: some of them are shorter, but not easy to find Here is a particularly elegant one, based on

16 Chapter 1 Some Useful Substitutions

the linearization technique: the inequality is homogeneous, so we may assume

that a + b + e = 1 We need to prove that

The point is to bound from below ( I-a +a (1-~t)2 2 by an affine function of a, suitably

chosen The best choice is the following

Of course, the question is how one came up with something like this Well, it

is actually very easy: we need constants A, B such that

for all a E [0, 1], with equality for a = ~ Impos~ng also th~ vanisfi~g of the

derivative of the difference between the left and nght- hand SIde at 3" YIelds the

desired constants A, B Once we have the previous estimates, it is very easy

to conclude by adding them and taking into account that a + b + e = 1

13 Find all real numbers k with the following property: for all positive

numbers a, b, e the following inequality holds

Vietnam TST 2009

Proof Take first of all a = b = 1 and arbitrary e to get that

1.2 The relations abc = a + b + e + 2 and ab + be + ea + 2abe = 1 17

Letting e -+ 0, we deduce that any k as in the statement must satisfy

-to reduce the problem -to

b y=e+a'

-is equivalent to L, a(b - e)2 2: o

In condusion, the solutions of the problem are the real numbers k such

18 Chapter 1 Some Useful Substitutions

We end this section with a very challenging problem, which answers the

following natural question: what would be the analogue of the classical

sub-stitution x = b!e, y = eta, Z = a~b when we have five variables? We warn the

reader that the first solution is really not natural

14 Let aI, a2, , a5 be positive real numbers such that

What is the least possible value of a\ + :2 + + a5

Gabriel Dospinescu, Mathematical Reflections

Proof Consider the linear system

We can try to solve this system by expressing, for example, all variables in

terms of Xl, X5 Namely, we can use the fifth, first and second equation to

express X2, X3, X4 in terms of Xl, X5 Replacing the obtained values in the

other two equations and eliminating Xl, x5 between them, we obtain that the

system has a nontrivial solution if and only if

II ai = 2 + L ai + ala4 + a4 a 2 + a2 a 5 + a5 a 3 + a3 a l·

All the previous (painful!) computations are left to the reader, since they

are far from having anything conceptual Note that the same result can be

obtained by computing the determinant of the associated matrix Now, the

previous relation is almost the one given in the statement, up to a permutation

of the variables So, since the conclusion is symmetric in the five variables, we

will assume from now on that the previous relation is satisfied instead of the

one given in the statement

1.2 The relations abc = a + b + e + 2 and ab + be + ea + 2abe = 1 19

The crucial claim is that under the previous hypothesis, the system has solutions whose unknowns Xi are positive Probably the easiest way to prove this is to exhibit such a solution Well, simply take Xl = 1,

1 + al + a2 + a3 + ala3 X5 = -= -= -=-:::

1 + a5 + a3a5 + a2 a 5

and the define

1 + X5 X4 + X5 X3 + X4 X4 = - - - , X3 = X2 = = ~

The question is, of course, how on earth did we choose the value of X5? Well, simply by solving the system, as indicated in the beginning of the solution Note that we clearly have Xi > 0 and an easy computation shows that these are solutions of the system (we only have to check two equations, since three are satisfied by construction)

The conclusion is that if the ai's satisfy the condition

20 Chapter 1 Some Useful Substitutions

this reduces to (~Xi)2 ::; 5 ~ x;, which is Cauchy-Schwarz

Putting everything together, we obtain that the minimal value of

is 5/2, attained when all ai's are equal to 2 Who wants to play now the same

Proof Here is another proof, also far from being evident " We will use the

following nontrivial

Lemma 1.4 For all nonnegative real numbers Xl, X2, ,X5 we have

(Xl + X2 + + X5)3 :2: 25(XIX2X3 + X2X3X4 + + X5XIX2)

Proof This is not easy at all, but here is a very elegant (but unnatural) proof:

consider the identity

XIX2X3 + X2X3X4 + X3X4X5 + X4X5XI + X5 X I X 2

= X5(XI + X3)(X2 + X4) + X2X3(XI + X4 X5)

We may assume that X5 = min Xi (since the inequality is cyclic), so that

Xl + X4 - X5 :2: O Denoting Xl + X2 + X3 + X4 = 4t and using the AM-GM

inequality, we deduce that

Thus, it remains to prove that

4t X5 + 3 ::; 25

By homogeneity, we may assume that X5 = 1 and then expanding everything

the inequality becomes, with the substitution 4t 1 = 3x, (x-1)2(8x+7):2: 0,

1.3 The relation a 2 + b 2 + c2 + 2abc = 1

Using this lemma for the inverses of the ai's and denoting

On the other hand, by Mac-Laurin's and AM-GM inequalities we also have

Taking into account these inequalities and the hypothesis, we deduce that

which immediately implies that S :2: ~ o

1.3 The relation a2 + b2 + c2 + 2abc = 1

The following problem is a rather tricky application of the AM-GM equality

in-15 Prove that in all acute-angled triangles the following inequality holds

( COS A) 2 (cos B) 2 (cos C) 2

cosB + cosC + cosA + 8cosAcosBcosC:2: 4

Titu Andreescu, MOSP 2000

Proof Since

cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 ,

22 Chapter 1 Some Useful Substitutions

the inequality can be written

(- -COSA)2 + (COSB)2 - C + (cosC)2 - A 2: 4(cos2 A+cos 2 B+cos 2 C)

Let

x = cos2 A, y = cos2 B, z = cos2 C,

so we need to prove that

The following problem is a preparation for a hard problem to come, but

has also an independent interest

16 Prove that in every acute-angled triangle ABC,

(cos A + cosB)2 + (cosB + cosC)2 + (cosC + cosA)2 :S 3

Proof Letting

cos A = 0,

YYz cosB = Y-;; 0, cosC = Y-;;Y 0

for some positive real x, y, z, we obtain the relation xyz = x + y + z + 2 from the classical identity L cos2 A + 2 n cos A = 1 Thus we can find positive real numbers a, b, c such that

which (after clearing denominators) is equivalent to

L (x 2 (y + z) + yzJ(x + z)(x + y)) :S ~(x + y)(y + z)(z + x)

and then to

1

L Jyz(x + y) yz(x + z) :S 3xyz + 2 LYz(y + z)

24 Chapter 1 Some Useful Substitutions

However, this follows immediately from the AM-GM inequality:

1

Jyz(x + y) yz(x + z) :S xyz + "2Yz(y + z)

Even though the following problem seems classical, it is actually rather

hard Fortunately, we did the difficult job in the previous problem We

also present an independent and very elegant approach due to Oaz Nair and

Richard Stong

17 Prove that if a, b, c 2 0 satisfy a2 + b2 + c2 + abc = 4, then

o :S ab + bc + ca - abc :S 2

Titu Andreescu, USAMO 2001

Proof The inequality on the left is easy: the hypothesis and the AM-GM

inequality imply that abc :S 1, so that min(a, b, c) :S 1 We may assume that

c = min(a, b, c), so that

The hard point is proving that ab + bc + ca - abc :S 2 Taking into account

the hypothesis, this can also be written as

( a ; b) 2 + (b; c) 2 + ( a ; c) 2 :S 3

If we denote a = 2x,b = 2y,c = 2z then x 2 + y2 + z2 + 2xyz = 1 and so

there exists a triangle ABC such that x = cos(A), y = cos(B), z = cos(C)

Thus, the problem reduces to the previous one (since a, b, c are nonnegative,

Proof Here is a very elegant proof for the hard part of the inequality Among

the three numbers a-I, b - 1, c - 1, two have the same sign, say b - 1 and c - 1

Thus (1 b)(1 - c) 2 0, implying that b + c :S bc + 1 and ab + ac - abc :S a

Thus, it is enough to prove that a + bc :S 2 But the given condition and the fact that a is nonnegative imply that

cos2 A = yz, cos2 B = zx, cos2 C = xy

Then xy + yz + zx + 2xyz = 1 and so there exist positive numbers X, Y, Z

26 Chapter 1 Some Useful Substitutions

On the other hand, we have

and it is a consequence of the following famous inequality

Lemma 1.5 For all positive numbers a, b, e we have

(ab + be + ea) (b + e)2 + (e + a)2 + (a + b)2 2': 4'

Proof We may assume that a 2': b 2': e First, we show that

-=-+ (a+b)2 (a+e)2 + (b+e)2 - 4ab >-+ (a+e)(b+e)

This can be rewritten

( 1 1)2 (a b)2

a + e - b + e 2': 4ab( a + b)2 ,

or equivalently 4ab(a+b)2 2': (a+e)2(b+e)2 This is clear, as (a+b)2 2': (a+e)2

and 4ab 2': (b + e)2

Thus, it remains to prove that

2(ab + be + ea) 2e2

We would like to thank the following people for their solutions: Vo Quoc

Ba Can (problems 13, 18), Xiangyi Huang (problems 4, 5), Logeswaran janugen (problem 1, 9), Oaz Nair (problem 17), Dusan Sobot (problems 2, 9, 16), Richard Stong (problems 8, 17), Gjergji Zaimi (problems 2, 3, 6, 7, 8, 11,

La-12, 15, 16, 17)

Chapter 2

Always Cauchy-Schwarz

As the title suggests, all problems in this chapter can be solved using the Cauchy-Schwarz inequality, even though sometimes this will require quite a lot of work Let us recall the statement of the Cauchy-Schwarz inequality: if

aI, a2,···, an and bl , b2, , bn are real numbers, then

This follows easily from the fact that L~=l (aix + bi )2 :2: 0 for all real numbers

x Indeed, the left-hand side is a quadratic function of x with nonnegative

values, so its discriminant is negative or O But this is precisely the content of the Cauchy-Schwarz inequality Another proof is based on Lagrange's identity

L (aibj - aj bi )2

I :Si<j:Sn

This useful identity will be used several times in this chapter

As the best way to get familiar with this inequality is via a lot of examples

of all levels of difficulty, we will not insist on any theoretical aspects and go directly to battle We start with two easy examples, destined to give the reader some confidence He will surely need it for the more difficult problems

to come

30 Chapter 2 Always Cauchy-Schwarz

1 Let a, b, c be nonnegative real numbers Prove that

for all nonnegative real numbers x

Titu Andreescu, Gazeta Matematica

Proof This is just a matter of re-arranging terms and applying

2 Let p be a polynomial with positive real coefficients Prove that if

p (t) 2 p(~) is true for x = 1, then it is true for all x > O

Titu Andreescu, Revista Matematica Timi§oara

Proof Write p(X) = ao + alX + + anxn and observe that

p(x)p ;: =(aO+aIX+···+anx) ao +-;:-+···+ xn

2 (ao + al + + an)2

= p(I)2

The following exercise is already a bit trickier, due to the lack of symmetry

3 Prove that for all real numbers a, b, c 2 1 the following inequality holds:

31

Proof The inequality being symmetric in b, c, but not in a, it is natural to

deal first with v1J=1 + JC=l This is easy to bound using Cauchy-Schwarz:

So, it is enough to prove that VbC + va=I :::; J a( bc + 1) But this is once

Another easy, but a bit exotic application of the Cauchy-Schwarz ity is the following Chinese olympiad problem

inequal-4 Let n be a positive integer Find the number of ordered n-tuples of integers (a I, a2, , an) such that

But then (again by Cauchy-Schwarz) we have ai + a§ + + a; 2 n 3 forcing

ai + a§ + + a; E {n3, n 3 + I} If ai + a~ + + a; = n 3, then we must have equality in Cauchy-Schwarz, implying that all ai's are equal to n Assume

now that ai + a§ + + a; = n 3 + 1 and let bi = ai - n Then

bI + b§ + + b~ = n 3 + 1 - 2n n 2 + n 3 = 1, forcing all but Olle bi vanish This is however impossible, as bl +b2+ -+bn = O Therefore, this second case will not occur and the only solution is

D

We continue the series of easy exercises:

32 Chapter 2 Always Cauchy-Schwarz

5 Let Xl, X2, , XlO be real numbers between 0 and ~ such that

1

2': "3(sinx2 + sinx3 + '" + sinxlO) and similarly for the other variables The result follows by adding up these

The following exercise combines an easy application of the

Cauchy-Schwarz inequality with some classical formulae from geometry Recall that r

is the inradius and that s is the semi-perimeter of a triangle

6 The triangle ABC satisfies

Show that ABC is similar to a triangle whose sides are integers, and

find the smallest set of such integers

Titu Andreescu, USAMO 2002

Proof Let the incircle be tangent to the sides of the triangle at points which

split the sides into segments of length X and y, X and z, and y and z Thus,

the sides of the triangle have lengths X + y, X + z, y + z

Thus, we are in the equality case of the Cauchy-Schwarz inequality, which is

precisely the case when there is a k > 0 such that X = 4k, z = 36k, y = 9k

Then the sides of the triangle are 13k, 40k, 45k Thus the triangle is similar

to the triangle with sidelengths 13,40,45, and the problem is solved D

The statement of the following problem looks rather classical There are however some technical problems which make the problem more difficult than expected

7 Let n 2': 2 be an even integer We consider all polynomials of the form

xn + an_Ix n - 1 + + alX + 1, with real coefficients and having at least one real zero Determine the least possible value of ai + a~ + + a~_l'

Czech-Polish-Slovak Competition 2002

34 Chapter 2 Always Cauchy-Schwarz

Proof Suppose that the corresponding polynomial has a real zero x Using

and so we need to find first the minimal value of f Now, looking at the

zeros of the derivative of f suggests that the minimal value might be taken at

x = 1, so that it equals n~l' This is not easy to prove using derivatives, as

the computations are a bit nasty Instead, we will prove in an elementary way

On the other hand, multiplying the last inequality by xn and adding the result

to the previous inequality gives the inequality

n ~ 2 (xn + 1)2 ~ x2 + + xn-2 + xn+2 + + x2(n-1)

Thus it remains to prove that (xn + 1)2 ~ 4xn , which is clear

The previous paragraph shows that

a1 + a2 + + an-l ~ - - 1

n-if xn + an_1Xn- 1 + + a1x + 1 has at least one real zero On the other hand,

choosing a1 = a2 = = an-1 = - n=-l shows that n~l is optimal D

35

The denominators in the following inequality look awful, but a clever application of the Cauchy-Schwarz inequality can make all of them equal The method of proof is worth remembering, since it appears quite often

8 Prove that for any positive real numbers x, y, z such that xyz ~ 1 the following inequality holds

Than Le, KaMaL magazine

Proof Using Cauchy-Schwarz, we obtain

' " ' 2 (x+y+z)2

L x ~ 3 ~ ijxyz( x + y + z) ~ x + y + z D

Here is a rather nice-looking inequality taken from a Romanian Team Selection Test There are plenty of ways to prove it, but what follows is particularly elegant and naturaL

36 Chapter 2 Always Cauchy-Schwarz

9 Let n 2: 2 and let aI, a2,· , an and bl, b2, , bn be real numbers such

that

ai + a§ + + a~ = bi + b§ + + b~= 1

and albl + a2b2 + " + anbn = O Prove that

Cezar and Tudorel Lupu, Romanian TST 2007

Proof The proof mimics the proof of Cauchy-Schwarz: take any real number

x and apply the Cauchy-Schwarz inequality to obtain

i)a~ + xbt )2 2: (L~-l a~ +: L~l bt )2 i=l

By hypothesis, the left-hand side equals 1 + x2 Therefore

(~ ai + x ~ b i r s *2 + 1)

for any real number x The difference between the right-hand side and the

left-hand side being a quadratic polynomial which takes nonnegative values

on the whole real line, its discriminant has to be negative or zero, thus

where A = L~=l ai and B = L~l b i It is immediate to check that this is

The following problem is easy, but the lack of symmetry might make it

appear more difficult than it really is

10 Find the largest real number T with the following property: if a, b, c, d, e

are nonnegative real numbers such that a + b = c + d + e, then

On the other hand,

Va + Vb::=; J2(a + b), Vc + vfi + y'e ::=; J3(c + d + e) = J3(a + b),

therefore

Combining these two inequalities yields the estimate

J a 2 + b 2 + c2 + d 2 + e2 > V30 ( Va + Vb + Vc + Vd + y'e)2

To see that this is optimal, it suffices to keep track of the equalities in the

previous inequalities For instance, we can take a = b = 3 and c = d = e = 2

38 Chapter 2 Always Cauchy-Schwarz

Proof Let ai = l+Xi' so that D ai = 1 and Xi = ai' Then, using

Cauchy-Schwarz we can write:

' " ' " Ja2 + a3 + + an ' " .fii2 + y'a3 + + Fn

Rearranging terms in the previous sum gives

1 '" y'al (_1_ + _1_ + + _1_)

and using again Cauchy-Schwarz we can bound this from below by

1 ' " (n-1)2y'al

vn=I' ~.fii2 + y'a3 + + Fn'

Using once more Cauchy-Schwarz for the denominators of each fraction

.fii2 + y'a3 + + Fn ::; J(n - 1)(a2 + a3 + + an)

The idea of the following example is worth keeping in mind, since it turns

out to be useful in a wide range of problems

12 For n ~ 2 let aI, a2, an be positive real numbers such that

(al + a2 + + an) (~+ ~ + +~) ::; (n + ~)2

Prove that max( aI, a2, ,an) ::; 4 min( aI, a2, , an)

Titu Andreescu, USAMO 2009

Proof The idea is to fix two of the variables, say aI, a2 and apply the

Cauchy-Schwarz inequality to get rid of the remaining variables Explicitly, this can

be written in the form

Let us write

ti,j = (If; - [~J 2

Now, expanding again gives us

40 Chapter 2 Always Cal1chy-Schwarz

This follows from Cauchy-Schwarz inequality, but there is still a detail to be

explained: we have to show that we cannot have equality But if we had

equality, all ti,j would be equal to n(n 2 _1) and so all numbers ~; + ~~ would be

equal Since n > 2, this would force an equality Xj = Xk for some j =I=- k But

We.continue with an inequality which combines a rather direct application

of the Cauchy-Schwarz inequality with a nice telescopic identity We also

present a beautiful alternate solution, due to Richard Stong

14 Prove that for any real numbers Xl, X2,· ,Xn the following inequality

holds:

1 + xi 1 + Xl + x 2 1 + Xl + + xn

Bogdan Enescu, IMO Shortlist 2001

Proof The solution is very short, but far from obvious The idea is to use the

Cauchy-Schwarz inequality to reduce the problem to

41

Thus, we need to prove that for any al,a2,' ,an ~ 0 we have

Define Si = a I + a2 + + ai, with the convention So = O This is an increasing sequence and so

Proof First note that for any nonnegative real numbers c and A we have

Here the first inequality is just 1 + A2 + x2 ~ 1 + A2 with equality if and only if X = 0; the second is Cauchy-Schwarz with equality if and only if

xVc = \11 + A2 Thus we cannot have equality in both cases and the final inequality is actually strict Applying this inequality repeatedly, it is easy to see by downwards induction on k that

42 Chapter 2 Always Cauchy-Schwarz

The trick of making the denominators equal thanks to a smart application

of Cauchy-Schwarz or AM-GM inequality is also used in the following problem

15 Prove that if a, b, c, d are positive real numbers, then

Inserting this in the previous inequality yields the desired result (note that

the inequality is strict, since otherwise the equality in the Cauchy-Schwarz

inequality would yield a = b = c = d, for which the inequality is strict) Note

that even though the inequalities we used were very rough, the constant 4 is

optimal: simply take a = band c, d close to O 0

One needs rather good gymnastics with Cauchy-Schwarz to deal with the

ak = V2k(:~~-k) and take Xo = Xn+l = 0 by convention, then equality occurs

if (-I)k- j Xj interpolates linearly and evenly between these endpoints and

Xk = ak· The explicit formula is

44 Chapter 2 Always Cauchy-Schwarz

Proof We see that we have equality when a = b = c, so we have to apply

the Cauchy-Schwarz inequality in a smart way for the left-hand side of the

inequality Namely, start with

This reduces the problem to showing that

Fortunately, this is equivalent to the classical inequality

and we are done

1 1 1 9

+ + >

D

The form of the following problem strongly suggests using

Cauchy-Schwarz However, it is rather easy to check that many attempts fail

18 Let x, y, z be real numbers and let A, B, C be the angles of a triangle

Prove that

xsinA + ysinB + zsinC::; J(1 + x 2)(1 + y2)(1 + z2)

Proof Using that C = 7f - A B, we obtain the identity

x sin A + ysinB + z sinC = (x + zcos B) sin A + zsinB cos A + ysinB

Using Cauchy-Schwarz and the fact that sin2 A + cos2 A = 1, we obtain

(x + z cos B) sin A + z sin B cos A ::; V (x + z cos B) 2 + z2 sin2 B

= Jx2 + z2 + 2xzcosB

Another application of Cauchy-Schwarz thus yields

x sin A + y sin B + z sin C ::; V(1 + y2)(x2 + z2 + 2xz cos B + sin2 B)

Thus, it remains to prove that

of Richard Stong, which makes things rather clear

19 Let a, b, c, x, y, z be real numbers and let

A = ax + by + cz, B = ay + bz + cx, C = az + bx + cy

Assuming that min(IA - BI, IB CI, IC - AI) 2': 1, find the smallest

possible value of (a 2 + b2 + c2)(x2 + y2 + z2)

Adrian Zahariuc, Mathematical Reflections

Proof· Note that by Lagrange's identity and the Cauchy-Schwarz inequality

(a 2 + b 2 + c 2 )(x2 + y2 + z2)

2': max (A2 + IB - CI2 B2 + IC - AI2 C 2 + IA B12)

46 Chapter 2 Always Cauchy-Schwarz

Using the hypothesis, it is immediate to check that the last quantity is at

least ~

To see that the answer of the problem is ~, it remains to find a 6-tuple

(a, b, c, x, y, z) satisfying the conditions of the problem and for which

Taking A = 1,B = O,C = -1 (this is a triple which minimizes

( A2 IB - CI

2

B2 IC - AI2 C 2 IA - B12)

max + 3 ' + 3 ' + 3

under the restrictions of the problem), we must ensure that we have equality

in the Cauchy-Schwarz inequality Solving the corresponding system yields,

after some tedious but easy work, suitable values for a, b, e, x, y, z, namely

Proof Let u be the column vector with entries (a, b, c), v the column vector

with entries (x, y, z) and let R be the linear map

Then A = uTv, B = uTRv, C = uT R 2 v and we want to minimize IIul1 2 ·llvI1 2

Note that R acts on ]R3 by fixing the line x = y = z and rotating 1200

in the orthogonal plane x + y + z = 0

Write u = UI + U2, where UI lies on the line x = y = z and U2 lies in the

plane x + y + z = ° and similarly for v Then

Thus the contributions of UI and VI to A, B, C are all the same (hence cancel

in IA - BI, etc.) Thus clearly the minimum occurs when UI = VI = 0 In

this case, R acts as a 1200

rotation on v, hence V + Rv + R 2 v = ° and thus

47

A + B + C = 0 Recall that cos a + cos(a + 21r /3) + cos(a + 41r /3) = ° for any

a If we now take w to be the angle between u and v, then we compute

A2 + B2 + C2 = IIuI12 '1IvI12(cos2 W + cos2(w + 21r/3) + cos2(w + 41r/3))

1

= 211uII2 '1IvI12(3 + cos2w + cos2(w + 21r/3) + cos2(w + 41r/3))

= ~llul12 ·llvI12

The conditions A + B + C = ° and min(IA - BI, IB - CI, IC - AI) 2 1

imply that A 2 + B2 + c2 2 2 (without loss of generality, assume that A and B

are nonnegative As IA - BI 2 1, we have max(A, B) 2 1, so A2 + B2 + C 2 =

2A2 + 2B2 + 2AB 2 2 max(A, B)2 2 2) and so

It is easy to see from the proof above that equality is attained Simply choose any (x,y,z) with x + y + z = ° and choose (a,b,e) with a + b + e = 0, orthogonal to (x, y, z) so that A = 0, and scaled so that B = 1 For example, taking (x,y,z) = (1, -2, 1) we see that (a,b,e) = (-1,0,1)/3 suffices 0

We present three short solutions for the following problem However, none

of them is really easy or natural

20 Let a, b, e, d be real numbers such that

(a 2 + 1)(b2 + 1)(e2 + 1)(d2 + 1) = 16

Prove that

-3::; ab + be + cd + da + ae + bd - abed::; 5

Titu Andreescu, Gabriel Dospinescu

Proof Write the inequality in the form

(ab+be+ed+da+ae+bd-abed-1)2::; 16

48 Chapter 2 Always Cauchy-Schwarz

Next, apply Cauchy-Schwarz in the form

(a(b + c + d - bed) + be + bd + cd - 1)2

::; (a2 + 1) [(b + c + d - bcd)2 + (be + bd + cd - 1)2]

The miracle is that we actually have

Of course, this can be checked by brute force, but perhaps nicer is determining

it through two applications of Lagrange's identity 0

Proof We will use Lagrange's identity and Cauchy-Schwarz in the form

and observe that the hypothesis becomes P(i)P( -i) = 16 This can also be

written as iP(iW = 16 On the other hand, we have

P(i) = 1 - L:: ab + abed + i (L:: a - L:: abc)

We deduce that

49

from where the conclusion follows, as the desired inequality can be written as

11 - L:: ab + abedl ::; 4

We continue with a really nice, but rather technical problem It quires some delicate algebraic computations and a rather exotic application

re-of Cauchy-Schwarz We also present a more conceptual prore-of, due to Richard Stong, which uses more advanced tools, but proves much more

21 Let a, b, c, d, e be nonnegative real numbers such that a2+b2+e2 = d2+e2

and a 4 + b 4 + c4 = d 4 + e 4 Prove that a 3 + b 3 + e3 ::; d 3 + e 3

IMC 2006

Proof Since we only deal with even exponents in the hypothesis, let us square

the desired inequality and write it in the form

Next, the identity

combined with the hypothesis easily yield

a 6 + b 6 + c6 = 3a 2 b 2 c 2 + d 6 + e 6

Thus, we need to prove that

With the obvious substitutions, this is equivalent to the inequality

50 Chapter 2 Always Cauchy-Schwarz

Using Cauchy-Schwarz

(22:: x3 + 3xyz)2 = (2:: x(2x2 + yz) f ::; (2:: x2) (~)2x2 + yz)2) ,

it remains to prove that

which follows immediately from x2y2 + y2 z2 + z2x2 2': xyz(x + y + z), itself

Proof Consider the polynomial p(t) = t3 - alt2 + a2 t - a3 with al and a2

fixed and a3 allowed to vary Regard the roots x, y, z of p( t) as functions of a3·

Suppose J : ffi -t ffi is any smooth function (at least three times differentiable

for the discussion below) Define a function

one merely checks that the curve

(

cs () = X+ (x-y)(x-z) ,y+ (y-z)(y-x) ,z+ (z-x)(z-y)

preserves aI, has b

vanishes at t = x, y, z Hence by two applications of Rolle's Theorem, its

second derivative vanishes at some <:; in (min(x, y, z), max(x, y, z)) and this is

The reader will probably appreciate the beauty of the following inequality

It is more difficult than it appears at first sight, as the obvious application of Cauchy-Schwarz fails rather badly

22 Prove that for any real numbers Xl, X2, ,Xn the following inequality holds

Thus, applying Cauchy-Schwarz yields

It is easy to compute the last expression and the final estimate that we obtain is

(

" I 0 0

1 ) 2 < 4n(n 2 - 1) ~ 2

52 Chapter 2 Always Cauchy-Schwarz

On the other hand, Legendre's identity shows that

i~l (Xi -x;)2 ~ 2n tx1-2 (tXi) 2

Well, unfortunately, when we combine all this we see that we are not

done, because of the bad term 2 CL:7=1 Xi)2 Fortunately, it is easy to repair

the argument: indeed, we can always add the same number to all xi'swithout

changing the hypothesis or the conclusion of the problem Thus, we may

assume that Xl + X2 + + Xn = O But then the previous inequalities allow

The following problem is a very tricky application of the Cauchy-Schwarz

inequality The technique used in the proof is worth remembering, since it is

quite useful It is also a standard tool in analysis and probability (it is actually

likely that the following problem is inspired by probability theory)

23 Let aI, a2, ,an be positive real numbers which add up to 1 Let ni be

the number of integers k such that 21-i 2: ak > 2-i Prove that

L Leindler, Miklos Schweitzer Competition

Proof Choose a positive integer N and split the sum in two parts: the one

for i ::; N and the one for i > N Apply the Cauchy-Schwarz inequality for

each of them to obtain

On the other hand, we have

Putting these inequalities together, we deduce that

Taking N = log2(n), we obtain an even stronger (and strict!) inequality, in

then any three of them are sides of a triangle

Adapted after IMO 2004

Proof The main point is to solve this problem for n = 3 If b, c are positive

real numbers, let us look at the possible values of

f(x) = (x + b + c) -( 1 + - + -1 1)

X b c when X 2: b + c It is not difficult to check (either directly or by computing the derivative) that f is increasing in this domain of x, so that

( 1 1 1 )

(b+c)2 f(x) 2: f(b + c) = 2(b + c) -b + - + - = 2 + 2 2: 10

54 Chapter 2 Always Cauchy-Schwarz

We deduce that if x, b, c satisfy f(x) < 10, then x, b, c are the sides of a triangle

(since everything is symmetric in x, b, c) Thus, for n = 3 the answer is 10

To reduce the general problem to the case n = 3, we uSG Cauchy-Schwarz,

which allows us to obtain information about Xl, X2, X3 knowing that

then any three among the numbers Xl, X2,"" Xn are the sides of a triangle

It remains to check that this is indeed optimal To see this, choose

X4 = X5 = = Xn = 1, ViO

Xl = X2 = -4-' Then Xl, X2, X3 are not the sides of a triangle and

25 If a, b, c, d, e are real numbers such that a + b + c + d + e = 0, then

Vasile Cartoaje

Proof First of all, we may assume that among a, b, c, d, e at least three of

the numbers are nonnegative, say a, b, c This follows immediately from the pigeonhole principle, possibly after changing the sign of all numbers The key step is the following very tricky application of Cauchy-Schwarz:

9(a 4 + b 4 + c 4 +d 4 + e 4 ) = [9(a 4 + b 4 + c 4 ) + 2(d 4 + e 4 )] + (7d4) + (7e4)

> -'-( 2-.:V_2_1 ,-( 9 -,-( a_ 4 _+_b 4 _+_c 4 -,-) _+_2-' ( d_ 4 _+_e_ 4 ; :) )_+_21 .:(_d2_+_e2-,-)~) 2

84 + 63 + 63

A small computation yields the equivalent form of this inequality

This reduces the problem to showing that

To exploit the relationship between a, b, c, d, e, we use the fact that

56 Chapter 2 Always Cauchy-Schwarz

Thus, it remains to prove that

for all nonnegative numbers a, b, c

This inequality does not seem to follow easily from well-known results, so

we will employ the powerful technique of mixing variables to prove it Let

f(a, b, c) = 36(a4 + b4 + c4) + (a + b + c)4 - 21(a2 + b2 + c2)2

= 15(a4 + b4 + c4) - 42(a2b2 + b2c2 + c2a2) + (a + b + c)4

We will first show that for a = min(a, b, c) we have

where the final inequality follows since 2Sb2 + 2Sc2 - 56a2 ~ O Thus we reduce

to the case where a ::; b = c In this case we compute

f(a, b, b) = 4(b4 + Sb3 - 15b2a2 + 2ba 3 + 4a4)

= 4(b - a)2(b2 + 10ab + 4a2)

~ O

The result follows

Note that we have equality for a = b = c = 2 and d = e = -3 o

57

Proof We will use a mixing variables argument Let

We want to prove that for a + b + c + d + e = 0 we have F( a, b, c, d, e) ~ O The basic formula we need is that

F(a, b, c, d, e) - F a, b, c, '

-2-= (d - e)2(21d2 + 34de + 21e2 - 7(a2 + b2 + c2)),

which can be checked by tedious computation

By the pigeonhole principle three of a, b, c, d, e must have the same sign (counting zero as having either sign) By symmetry, we may assume a, b, c ~ O Then

7(a2 + b2 + c2) ::; 7(a + b + c)2 = 7(d + e)2

::; 17(d + e)2 + 4(d2 + e2) = 21d2 + 34de + 21e2

Therefore by the formula above F( a, b, c, d, e) ~ F (a, b, c, d!e, d!e) Thus we may assume three of a, b, c, d, e are nonnegative and the other two are equal To keep the same basic formula, we invoke symmetry and switch our assumption

to c, d, e ~ 0 and a = b = -(c + d + e)/2 Now suppose c::; e Then we have

58 Chapter 2 Always Cauchy-Schwarz

We end this chapter with a challenging inequality, which combines some

clever uses of Cauchy-Schwarz with a tricky homogeneity argument This

result also appears in [35] and it is a generalization of a problem discussed in

[3], chapter 2, example 11

26 Prove that for any positive real numbers aI, a2, ,an, XI, X2,···, Xn

such that

the following inequality holds

Vasile Cartoaje, Gabriel Dospinescu

Proof First of all, it is enough to prove that for any Xl, X2, , Xn > 0 and

any aI, 0;2, , an > 0 we have

LI<i<j<nXiXj

G)

""' a I (X2 + + Xn) ~ n

~a2+···+an

Since this is homogeneous, it is enough to prove it when Xl + X2 + + Xn = 1

In this case, it is equivalent to

Ll<i<j<n XiXj

G)

But Cauchy-Schwarz shows that

and a second application of Cauchy-Schwarz yields

A very elegant approach for this inequality was proposed by Darij Grinberg

in [35], where a more general result is proved We may assume that

We need to prove that

Using T2's lemma (a form of Cauchy-Schwarz), we reduce this to proving that

60 Chapter 2 Always Cauchy-Schwarz

This follows directly from Cauchy-Schwarz and the identity

by mixing variables: consider the map

and set x = al !a2 We claim that F(al, a2,"" an) 2: F(x, x, a3,'" ,an),

A small computation shows that this is equivalent to

Another computation shows that

But this is easy, since the left-hand side is at least 2 2.:i>3 ai = 2(1 - 2x)

and 2(1 - 2x) 2: li'=-~x is equivalent to (1 - 2x)2 2: 0.- Thus, we have

F(al' a2,··· ,an) 2: F(x, x, a3,· , an) Continuing to mix variables in this

way and using the continuity of F implies that F(al' a2, ,an) is at least

F(m, m, , m), where m is the arithmetic mean of the ai's, namely lin The result follows

The following people provided solutions to the problems discussed in this chapter: Vo Quoc Ba Can (problem 20), Ta Minh Hoang (problem 17), Mitchell Lee (problem 6), Dung Tran Nam (problem 25), Dusan Sobot (prob-lems 1, 2, 5, 7), Richard Stong (problems 14, 19, 21, 25), Gjergji Zaimi (prob-lems 3, 4, 10, 13)