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Basics of Olympiad Inequalities

Samin Riasat

ii

Introduction

The aim of this note is to acquaint students, who want to participate in mathematical Olympiads, to

Olympiad level inequalities from the basics. Inequalities are used in all elds of mathematics. They have

some very interesting properties and numerous applications. Inequalities are often hard to solve, and it is

not always possible to nd a nice solution. But it is worth approaching an inequality rather than solving

it. Most inequalities need to be transformed into a suitable form by algebraic means before applying some theorem. This is what makes the problem rather dicult. Throughout this little note you will nd dierent ways and approaches to solve an inequality. Most of the problems are recent and thus need a fruitful combination of wisely applied techniques. It took me around two years to complete this; although I didn't work on it for some months during

this period. I have tried to demonstrate how one can use the classical inequalities through dierent ex-

amples that show dierent ways of applying them. After almost each section there are some exercise

problems for the reader to get his/her hands dirty! And at the end of each chapter some harder problems

are given for those looking for challenges. Some additional exercises are given at the end of the book for

the reader to practice his/her skills. Solutions to some selected problems are given in the last chapter to

present dierent strategies and techniques of solving inequality problems. In conclusion, I have tried to

explain that inequalities can be overcome through practice and more practice.

Finally, though this note is aimed for students participating in the Bangladesh Mathematical Olympiad

who will be hoping to be in the Bangladesh IMO team I hope it will be useful for everyone. I am really

grateful to the MathLinks forum for supplying me with the huge collection of problems.

Samin Riasat

28 November, 2008

iii ivINTRODUCTION

Contents

Introductioniii

1 The AM-GM Inequality 1

1.1 General AM-GM Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Weighted AM-GM Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 More Challenging Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Cauchy-Schwarz and H

older's Inequalities 9

2.1 Cauchy-Schwarz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 H

older's Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3 More Challenging Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Rearrangement and Chebyshev's Inequalities 19

3.1 Rearrangement Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 Chebyshev's inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.3 More Chellenging Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Other Useful Strategies 27

4.1 Schur's Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.2 Jensen's Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3 Minkowski's Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.4 Ravi Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.5 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.6 Homogenization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5 Supplementary Problems 31

6 Hints and Solutions to Selected Problems 33

References39

v viCONTENTS

Chapter 1

The AM-GM Inequality

1.1 General AM-GM Inequality

The most well-known and frequently used inequality is the Arithmetic mean-Geometric mean inequality or widely known as the AM-GM inequality. The term AM-GM is the combination of the two terms Arithmetic MeanandGeometric Mean. The arithmetic mean of two numbersaandbis dened bya+b2 Similarlypabis the geometric mean ofaandb. The simplest form of the AM-GM inequality is the following: Basic AM-GM Inequality.For positive real numbersa;b a+b2 pab:

The proof is simple. Squaring, this becomes

(a+b)24ab; which is equivalent to (ab)20: This is obviously true. Equality holds if and only ifa=b.

Example 1.1.1.For real numbersa;b;cprove that

a

2+b2+c2ab+bc+ca:

First Solution.By AM-GM inequality, we have

a

2+b22ab;

b

2+c22bc;

c

2+a22ca:

Adding the three inequalities and then dividing by 2 we get the desired result. Equality holds if and only

ifa=b=c.

Second Solution.The inequality is equivalent to

(ab)2+ (bc)2+ (ca)20; 1

2CHAPTER 1. THE AM-GM INEQUALITY

which is obviously true. However, the general AM-GM inequality is also true for anynpositive numbers. General AM-GM Inequality.For positive real numbersa1;a2;:::;anthe following inequality holds. a

1+a2++ann

npa

1a2an;

with equality if and only ifa1=a2==an. Proof.Here we present the well known Cauchy's proof by induction. This special kind of induction is done by performing the following steps: i. Base case. ii.Pn=)P2n. iii.Pn=)Pn1. HerePnis the statement that the AM-GM is true fornvariables. Step 1: We already proved the inequality forn= 2. Forn= 3 we get the following inequality: a+b+c3

3pabc:

Lettinga=x3;b=y3;c=z3we equivalently get

x

3+y3+z33xyz0:

This is true by Example 1.1.1 and the identity

x

3+y3+z33xyz= (x+y+z)(x2+y2+z2xyyzzx):

Equality holds forx=y=z, that is,a=b=c.

Step 2: Assuming thatPnis true, we have

a

1+a2++ann

npa

1a2an:

Now it's not dicult to notice that

a

1+a2++a2nnnpa

1a2an+nnpa

n+1an+2a2n2n2npa

1a2a2n

implyingP2nis true.

Step 3: First we assume thatPnis true i.e.

a

1+a2++ann

npa

1a2an:

As this is true for all positiveais, we letan=n1pa

1a2an1. So now we have

a

1+a2++ann

nqa

1a2an1n1pa

1a2an1

nq(a1a2an1)nn1 n1pa

1a2an1

=an;

1.1. GENERAL AM-GM INEQUALITY3

which in turn is equivalent to a

1+a2++an1n1an=n1pa

1a2an1:

The proof is thus complete. It also follows by the induction that equality holds fora1=a2==an. Try to understand yourself why this induction works. It can be useful sometimes. Example 1.1.2.Leta1;a2;:::;anbe positive real numbers such thata1a2an= 1. Prove that (1 +a1)(1 +a2)(1 +an)2n:

Solution.By AM-GM,

1 +a12pa

1;

1 +a22pa

2;

1 +an2pa

n: Multiplying the above inequalities and using the facta1a2an=1 we get our desired result. Equality holds forai= 1; i= 1;2;:::;n. Example 1.1.3.Leta;b;cbe nonnegative real numbers. Prove that (a+b)(b+c)(c+a)8abc:

Solution.The inequality is equivalent to

a+bpab b+cpbc c+apca 222;
true by AM-GM. Equality holds if and only ifa=b=c.

Example 1.1.4.Leta;b;c >0. Prove that

a 3bc +b3ca +c3ab a+b+c:

Solution.By AM-GM we deduce that

a 3bc +b+c33ra 3bc bc= 3a; b 3ca +c+a33rb 3ca ca= 3b; c 3ab +a+b33rc 3ab ab= 3c:

4CHAPTER 1. THE AM-GM INEQUALITY

Adding the three inequalities we get

a 3bc +b3ca +c3ab + 2(a+b+c)3(a+b+c); which was what we wanted. Example 1.1.5. (Samin Riasat)Leta;b;cbe positive real numbers. Prove that ab(a+b) +bc(b+c) +ca(c+a)X cycabra b (b+c)(c+a):

Solution.By AM-GM,

2ab(a+b) + 2ac(a+c) + 2bc(b+c)

=ab(a+b) +ac(a+c) +bc(b+c) +ab(a+b) +ac(a+c) +bc(b+c) =a2(b+c) +b2(a+c) +c2(a+b) + (a2b+b2c+a2c) + (ab2+bc2+a2c) a2(b+c) +b2(a+c) +c2(a+b) + (a2b+b2c+a2c) + 3abc =a2(b+c) +b2(a+c) +c2(a+b) +ab(a+c) +bc(a+b) +ac(b+c) =a2(b+c) +ab(a+c)+b2(a+c) +bc(a+b)+c2(a+b) +ac(b+c) 2pa

3b(b+c)(a+c) + 2pb

3c(a+c)(a+b) + 2pc

3a(a+b)(b+c)

=2abra b (b+c)(a+c) + 2cbrb c (a+c)(a+b) + 2acrc a (a+b)(b+c):

Equality holds if and only ifa=b=c.

Exercise 1.1.1.Leta;b >0. Prove that

ab +ba 2: Exercise 1.1.2.For all real numbersa;b;cprove the following chain inequality

3(a2+b2+c2)(a+b+c)23(ab+bc+ca):

Exercise 1.1.3.Leta;b;cbe positive real numbers. Prove that a

3+b3+c3a2b+b2c+c2a:

Exercise 1.1.4.Leta;b;cbe positive real numbers. Prove that a

3+b3+c3+ab2+bc2+ca22(a2b+b2c+c2a):

Exercise 1.1.5.Leta;b;cbe positive real numbers such thatabc= 1. Prove that a

2+b2+c2a+b+c:

1.2. WEIGHTED AM-GM INEQUALITY5

Exercise 1.1.6. (a)Leta;b;c >0. Show that

(a+b+c)1a +1b +1c 9: (b)For positive real numbersa1;a2;:::;anprove that (a1+a2++an)1a 1+1a 2++1a n n2: Exercise 1.1.7.Leta;b;cbe nonnegative real numbers such thata+b+c= 3. Prove that a

2+b2+c2+ab+bc+ca6:

Exercise 1.1.8.Leta;b;c;d >0. Prove that

a 2b +b2c +c2d +d2a a+b+c+d:

1.2 Weighted AM-GM Inequality

The weighted version of the AM-GM inequality follows from the original AM-GM inequality. Suppose thata1;a2;:::;anare positive real numbers and letm1;m2;:::;mnbe positive integers. Then we have by AM-GM, a

1+a1++a1|{z}

m

1+a2+a2++a2|{z}

m

2++an+an++an|{z}

m nm

1+m2++mn

0 a1a1:::a1|{z} m 1a

2a2:::a2|{z}

m

2anan:::an|{z}

m n1 A1m

1+m2++mn

This can be written as

m

1a1+m2a2++mnanm

1+m2++mn(am11am22amnn)1m

1+m2++mn:

Or equivalently in symbols

PmiaiP

miYamii 1P mi:

Lettingik=mkP

mj=mkm

1+m2++mnfork= 1;2;:::;nwe can rewrite this as follows:

Weighted AM-GM Inequality.For positive real numbersa1;a2;:::;anandnweightsi1;i2;:::;in such thatnX k=1i k= 1, we have a

1i1+a2i2++aninai11ai22ainn:

6CHAPTER 1. THE AM-GM INEQUALITY

Although we have a proof ifi1;i2;:::;inare rational, this inequality is also true if they are positive real

numbers. The proof, however, is beyond the scope of this note. Example 1.2.1.Leta;b;cbe positive real numbers such thata+b+c= 3. Show that a bbcca1

Solution.Notice that

1 = a+b+c3 ab+bc+caa+b+c abbcca

1a+b+c;

which impliesabbcca1. Example 1.2.2. (Nguyen Manh Dung)Leta;b;c >0 such thata+b+c= 1. Prove that a abbcc+abbcca+acbacb1:

Solution.From weighted AM-GM, we have

a

Summing up the three inequalities we get

(a+b+c)2aabbcc+abbcca+bacbac:

That is,

a abbcc+abbcca+acbacb1: Very few inequalities can be solved using only the weighted AM-GM inequality. So no exercise in this section.

1.3. MORE CHALLENGING PROBLEMS7

1.3 More Challenging Problems

Exercise 1.3.1.Leta;b;cbe positive real numbers such thatabc= 1. Prove that ab +bc +ca a+b+c: Exercise 1.3.2. (Michael Rozenberg)Leta;b;canddbe non-negative numbers such thata+b+c+d=

4. Prove that4abcd

ab +bc +cd +da Exercise 1.3.3. (Samin Riasat)Leta;b;cbe positive real numbers. Prove that a

3+b3+c33

a+b+c3 a2+b2+c23 a2b+b2c+c2a3 Exercise 1.3.4.(a) (Pham Kim Hung)Leta;b;cbe positive real numbers. Prove that ab +bc +ca +33pabc
a+b+c4: (b) (Samin Riasat)For real numbersa;b;c >0 andn3 prove that ab +bc +ca +n

33pabc

a+b+c! 3 +n: Exercise 1.3.5. (Samin Riasat)Leta;b;cbe positive real numbers such thata+b+c=ab+bc+ca andn3. Prove that a2b +b2c +c2a +3na

2+b2+c23 +n:

8CHAPTER 1. THE AM-GM INEQUALITY

Chapter 2

Cauchy-Schwarz and H

older's

Inequalities

2.1 Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is a very powerful inequality. It is very useful in proving both cyclic and

symmetric inequalities. The special equality case also makes it exceptional. The inequality states: Cauchy-Schwarz Inequality.For any real numbersa1;a2;:::;anandb1;b2;:::;bnthe following in- equality holds. with equality if the sequences are proportional. That is if a1b 1=a2b

2==anb

n. First proof.This is the classical proof of Cauchy-Schwarz inequality. Consider the quadratic f(x) =nX i=1(aixbi)2=x2nX i=1a 2ixnX i=12aibi+nX i=1b

2i=Ax2+Bx+C:

Clearlyf(x)0 for all realx. Hence ifDis the discriminant off, we must haveD0. This implies B 24AC)
nX i=12aibi! 2 4 nX i=1a 2i! nXquotesdbs_dbs24.pdfusesText_30

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