Strategies to Solve Olympiad-style Problems-II

Aditya Ghosh, India

July 24, 2017

This is the second chapter of my series of articles on strategies to solve olympiad-style problems1. In this article, we shall focus on inequalities. If a problem here isleft without a solution, then it is meant to be an exercise to the reader.

1 Some Basic Groundings

1.1 Introduction

The set of real numbers R is equipped with an ordering such that :

#1 For every x ∈ R, one and exactly one of these holds: x < 0, x = 0, x > 0.#2 We say x > y if and only if x− y > 0. And x ≥ y if either x > y or x = y.#3 If x ≥ 0, y ≥ 0 then x+ y ≥ 0 and xy ≥ 0 with equality iff x = y = 0.

It is very interesting that just using the above three, one can prove the follwingfacts: 1. x > 0 ⇐⇒ (−x) < 0. 2. If a > 0 then 1/a > 0.

3. If 0 < a < 1 then a2 < a. Else, a2 ≥ a.4. x ≥ y ⇒ x+ z ≥ y + z, ∀z ∈ R. And x ≥ y and a ≥ b⇒ x+ a ≥ y + b.5. x ≥ y and a > 0 ⇒ xa ≥ ya. For a < 0 we have xa ≤ ya.6. x ≥ y and a ≥ b ⇒ xa ≥ yb when either x, b ∈ R+ or a, y ∈ R+.7. If x ≥ y and when xy > 0, then 1

x≤ 1

y.

8. If a2 ≥ b2 and a + b > 0 then a ≥ b. (Note that, a2 ≥ b2 and a + b ≥ 0 neednot imply a ≥ b. For counterexample, take a = −1, b = 1.)

9. If a2 ≥ b2 and a ≥ 0 then a ≥ b.

How to prove them? some are solved below. Others are left as an exercise to you!1. If x > 0 and (−x) ≥ 0 then, using #3 we get 0 = x+ (−x) > 0, contradiction.Prove (−x) < 0⇒ x > 0 similarly.2. Suppose for some a > 0 but 1

a< 0. Then, using 1. we get − 1

a> 0. Now, using

#3, we get a · (− 1a) > 0⇒ −1 > 0, contradiction.

6. If x, b ∈ R+, then, xa− yb = x(a− b) + b(x− y) ≥ 0.(Using #3) Else, a, y ∈ R+,then, xa− yb = (x− y)a+ (a− b)y ≥ 0. (Alternately, one can use 5 twice.)

Another important consequence is that, x2 ≥ 0 for all real numbers x. Equalityholds here if and only if x = 0. Solution: Using #1, either x ≥ 0, then by #3, wehave x2 = x · x ≥ 0. Else, x < 0, and by 1. we have (−x) > 0. Hence by #3, wehave x2 = (−x) · (−x) > 0. Thus proved.

1All chapters available at https://mathsupportweb.wordpress.com/blog/

1

This inequality produces the following basic inequalities :

◦ (a− b)2 ≥ 0⇒ a2 + b2 ≥ 2ab for all a, b ∈ R.

◦ a2 + b2 ≥ 2ab⇒ ab

+ ba≥ 2 for all a, b such that ab > 0.

◦ Hence, for all a, b > 0 we have 1a

+ 1b≥ 4

a+b.

◦ 2(a2 + b2) = (a + b)2 + (a− b)2 ≥ (a + b)2 for all a, b ∈ R.

◦ (a− b)2 + (b− c)2 + (c− a)2 ≥ 0⇒ a2 + b2 + c2 ≥ ab+ bc+ ca ∀a, b, c ∈ R.

◦ Hence, 3(ab+bc+ca) ≤ (a+b+c)2 ≤ 3(a2+b2+c2). [why? expand (a+b+c)2.]

◦ Also, a3 + b3 + c3 − 3abc = (a+ b+ c)(a2 + b2 + c2 − ab− bc− ca) ≥ 0 for alla, b, c such that a+ b+ c ≥ 0.

The reader is requested to note the equality cases in each of the inequalitiesstated above. Next, we move to some easy problems to see how all these works.

• Prove that, a4 + b4 + c4 ≥ abc(a+ b+ c) for all a, b, c ∈ R.

Solution: Using x2 + y2 + z2 ≥ xy + yz + zx twice, we get

a4 + b4 + c4 ≥ (ab)2 + (bc)2 + (ca)2 ≥ ab · bc+ bc · ca+ ca · ab = abc(a+ b+ c).

• For a, b, c > 1 prove that, abc+ 1/a+ 1/b+ 1/c ≥ a+ b+ c+ 1/abc.

• If m,n ∈ Z+, show that, mn<√

2 if and only if√

2 < m+2nm+n

.

• For a, b, c ∈ R satisfying a4 + b4 + c4 + d4 = 16. Prove that, a5 + b5 + c5 + d5 ≤ 32.

• For a, b, c, d ∈ R, such that, a+ b > 0, c+ d > 0. Prove that,

ab

a+ b+

cd

c+ d≤ (a+ c)(b+ d)

a+ b+ c+ d.

Solution: It is equivalent to

ab+ab(c+ d)

a+ b+cd(a+ b)

c+ d+ cd ≤ (a+ c)(b+ d) = ab+ ad+ bc+ cd

or, ab(c+ d)2 + cd(a+ b)2 ≤ (ad+ bc)(a+ b)(c+ d)

or, ab(c2 + d2) + cd(a2 + b2) + 4abcd ≤ (ad+ bc)(ac+ bd) + (ad+ bc)2

or, 4abcd ≤ (ad+ bc)2 or, 0 ≤ (ad− bc)2.• Let a, b, c, d be real numbers satisfying a+ d = b+ c. Then Prove that,(a− b)(c− d) + (a− c)(b− d) ≥ (a− d)(b− c).• For a, b, c ∈ R, prove that |a|+ |b|+ |c| − |b+ c| − |c+ a| − |a+ b|+ |a+ b+ c| ≥ 0.

• Let x, y, z ≥ 0. Prove that x2 + xy2 + xyz2 ≥ 4(xyz − 1).

Solution: We use a+ b ≥ 4√ab three times:

4 + x2 + xy2 + xyz2 ≥ 4x+ xy2 + xyz2 ≥ 4xy + xyz2 ≥ 4xyz.

• Let x, y, z > 0. Prove that if xyz ≥ xy + yz + zx, then (a) xyz ≥ 3(x+ y + z),(b) x+ y + z ≥ 9.

Solution: (a) x2y2z2 ≥ (xy + yz + zx)2 ≥ 3xyz(x+ y + z), done!(b)1 (∑

cyc

xy)2≥ 3xyz

(∑cyc

x)≥ 3

(∑cyc

xy)(∑

cyc

x)⇒ xy + yz + zx ≥ 3(x+ y + z)

Hence, (x+ y + z)2 ≥ 3(xy + yz + zx) ≥ 9(x+ y + z).

1Notation :∑

cyc f(a, b, c) = f(a, b, c) + f(b, c, a) + f(c, a, b).

2

• If a, b, c, d > 0, then show that at least one of the following inequalities is wrong:

a+ b < c+ d, (a+ b)(c+ d) < ab+ cd, (a+ b)cd < ab(c+ d).

Solution: Assume that all of them are true. Multiplying the first two, we getab+ cd > (a+ b)2 ≥ 4ab⇒ cd ≥ 3ab . . . (1). Multiplying the last two, we getab(ab+ cd) > cd(a+ b)2 ≥ cd · 4ab⇒ ab > 3cd . . . (2). Adding (1) and (2) we getab+ cd > 3(ab+ cd)−not possible for a, b, c, d > 0.

• Let a, b, c > 0 with a+ b+ c = 1. If a+ b+ c > 1/a+ 1/b+ 1/c, then show thatexactly one among a, b, c is greater than 1.

• For any n ∈ N, prove that,

1

2≤ 1

n+ 1+

1

n+ 2+ . . .+

1

2n≤ 3

4.

Solution:1

n+ 1+

1

n+ 2+ . . .+

1

2n≥ 1

n+ n+

1

n+ n+ . . .+

1

2n=

n

2n.

1

n+

1

n+ 1+

1

n+ 2+ . . .+

1

2n=

1

2

[( 1

n+

1

2n

)+( 1

n+ 1+

1

2n− 1

)+ . . .+

( 1

2n+

1

n

)]

=3n

2

[1

2n2+

1

2n2 + (n− 1)+ . . .+

1

2n2

]≤ 3n

2

[(n+ 1)

1

2n2

]=

3

4+

1

n.

• Prove that, for any positive integer n, the fractional part of√

4n2 + n is smallerthan 1/4.

1.2 Standard ToolsHere is a list of standard tools which the reader can study from any standard book2:◦ Theory of quadratic functions◦ RMS ≥ AM ≥ GM ≥ HM◦ Cauchy-schwarz inequality/ Titu’s Lemma◦ Rearrangement inequality and Tchebyshev’s inequality◦ Weighted AM-GM inequality, Generalised Mean inequality◦ Convexity, Jensen’s inequality◦ Triangle inequality, Induction etc.Note: When you are applying Rearrangement ineq. or Tchebyshev’s ineq., you

should prove the ordering of the variables. And when applying ‘Weighted AM-GM’inequality, specify the weights properly.

The following tools are a bit advanced. Contestants should try to avoid them,atleast upto INMO :◦ Holder’s inequality, Minkowski’s inequality◦ Karamata inequality ◦ Schur’s inequality◦ u-v-w technique, SOS method, SMV method◦ Muirhead’s inequality ◦ Lagrange Multipliers

2You can study these from ‘An Excursion In Mathematics’.

3

1.3 Application of AM-GM

All a, b, c, d, x, y, z in this subsection stand for positive real numbers, unlessotherwise specified.• “By AM-GM, a2 +b2 ≥ 2ab. Similarly, b2 +c2 ≥ 2bc , c2 +a2 ≥ 2ca. Adding up

these inequalities altogether, we get a2+b2+c2 ≥ ab+bc+ca.”- shall be abbreviatedhere1 by saying∑

cyc

[a2 + b2 ≥ 2ab

]⇒ a2 + b2 + c2 ≥ ab+ bc+ ca.

In this way, we get the following inequalities :

∑cyc

[bc

a+ca

b≥ 2c

]⇒ bc

a+ca

b+ab

c≥ a+ b+ c

∑cyc

[a2

b2+b2

c2≥ 2

a

c

]⇒ a2

b2+b2

c2+c2

a2≥ a

c+b

a+c

b∑cyc

[1

b3+

1

c3+

1

d3≥ 3

1

bcd

]⇒ 1

a3+

1

b3+

1

c3+

1

d3≥ a+ b+ c+ d

abcd∑cyc

[a

b+a

b+b

c≥ 3

a3√abc

]⇒ a

b+b

c+c

a≥ a+ b+ c

3√abc∑

cyc

[a

b+b

c+b

c≥ 3

3√abc

a

]⇒ a

b+b

c+c

a≥ 3√abc

(1

a+

1

b+

1

c

)∑cyc

[a3b+ abc2 ≥ 2a2bc

]⇒ a3b+ b3c+ c3a ≥ abc(a+ b+ c).

• Prove that, (1 +

x

y

)(1 +

y

z

)(1 +

z

x

)≥ 2 +

x+ y + z3√xyz

.

• Prove that,√a2 + b2 + c2 + d2

4≥ 3

√abc+ bcd+ cda+ dab

4.

Solution:16(abc+ bcd+ cda+ dab) = 4(4ab(c+ d) + 4cd(a+ b))

≤ 4(a+ b)2(c+ d) + 4(c+ d)2(a+ b)

= 4(a+ b)(c+ d)(a+ b+ c+ d) ≤ (a+ b+ c+ d)3.

Hence,3

√abc+ bcd+ cda+ dab

4≤ a+ b+ c+ d

4≤√a2 + b2 + c2 + d2

4.

• Prove that, a8 + b8 + c8

a3b3c3≥ 1

a+

1

b+

1

c.

Solution 1: Using x2 + y2 + z2 ≥ xy + yz + zx multiple times:

a8 + b8 + c8 ≥ (ab)4 + (bc)4 + (ca)4 ≥ a2b2c2(a2 + b2 + c2) ≥ a2b2c2(ab+ bc+ ca).

Solution 2: Using ‘Generalised Mean Inequality’, namely M8 ≥M1.

a8 + b8 + c8

3≥ (a+ b+ c)2

32

(a+ b+ c

3

)6

≥ ab+ bc+ ca

3(abc)2.

1caution: Its not a standard notation, so don’t use it without mentioning the meaning of it.

4

• For 0 < a, b, c < 1, prove that,√abc+

√(1− a)(1− b)(1− c) < 1.

Solution:

L.H.S. <√bc+

√(1− b)(1− c) ≤ b+ c

2+

1− b+ 1− c2

= 1.

• Let a, b, c ∈ R+ such that a+ b+ c = 3. Prove that,√a+√b+√c ≥ ab+ bc+ ca.

Solution:∑cyc

[a2 +

√a+√a

AM−GM≥ 3a

]⇒ 9− 2

∑cyc

ab+ 2∑cyc

√a ≥ 3

∑cyc

a = 9.

• Let x, y, z ∈ R+ such that xyz = 1. Prove that,

x2 + y2 + z2 + xy + yz + zx ≥ 2(√x+√y +√z).

• Let a, b, c ∈ R+ such that abc = 1. Prove that,

1 + ab

1 + c+

1 + bc

1 + a+

1 + ca

1 + b≥ 3.

?? While applying AM-GM, you should carefully note the equality cases. Thefollowing example illustrates this.• Let x, y, z ∈ R+ such that xyz = 1. Prove that,

x3

(1 + y)(1 + z)+

y3

(1 + z)(1 + x)+

z3

(1 + x)(1 + y)≥ 3

4.

Solution: If we apply AM-GM in this way:

x3

(1 + y)(1 + z)+ (1 + y) + (1 + z) ≥ 3x⇒

∑cyc

x3

(1 + y)(1 + z)≥ x+ y + z − 6,

then we clearly fail to prove what we want. How to repair it? Note that, forx = y = z = 1, equality holds. And in that case, the x3

(1+y)(1+z)terms equal to 1/4.

So we should do like this:

x3

(1 + y)(1 + z)+

1 + y

8+

1 + z

8≥ 3x

4⇒∑cyc

x3

(1 + y)(1 + z)≥ x+ y + z

2− 3

4≥ 3

4.

where the last inequality is due to x+ y + z ≥ 3 3√xyz = 3.

Moral of the story: always apply AM-GM in a way that equality occurs.

• Let a, b, c ∈ R+. Prove that, (1 + a)(a+ b)(b+ c)(c+ 16) ≥ 81abc.

5

1.3.1 Weighted AM-GM-HM

• Prove that,a8 + b8 + c8

a3b3c3≥ 1

a+

1

b+

1

c.

Solution 3:∑cyc

[2a8 + 3b8 + 3c8

8≥ 8a2b3c3

]⇒ a8 + b8 + c8 ≥ a3b3c3

(1

a+

1

b+

1

c

).

• If a, b, c are positive real numbers such that abc = 1, prove that

ab+cbc+aca+b ≤ 1.Solution: a+ b+ c

3≥ 3√abc = abc⇒ 1 ≥ 3abc

a+ b+ cNow, by weighted AM-GM, (a, b, c as weights)

a+b+c

√(ab)c(bc)a(ab)c ≤ c · ab+ a · bc+ b · ca

a+ b+ c≤ 1.

Hence, ab+cbc+aca+b = (ab)c(bc)a(ab)c ≤ 1.

• For a fixed positive integer n, find the minimum value of the sum

x1 +x222

+x333

+ . . .+xnnn

given that x1, x2, . . . , xn are positive numbers satisfying that the sum of theirreciprocals is n.

Solution 1: By weighted AM-GM with 1 and k − 1 as weights,

1 · xkk + (k − 1) · 1 ≥ kxk ⇒xk

k+ 1− 1

k≥ xk

Summing up such inequalities for k = 1, 2, . . . , n we getn∑

k=1

xkkk

+ n−n∑

k=1

1

k≥

n∑k=1

xkAM−HM≥ n2∑

1/xk= n.

Hence,x1 +

x222

+x333

+ . . .+xnnn≥ 1 +

1

2+ . . .+

1

n.

With equality when x1 = x2 = . . . = xn = 1.

Solution 2: Let Hn = 1 + 1/2 + . . .+ 1/n. Takng 1k

as the weight for xk, we getn∑

k=1

1

k· xkk ≥ Hn(x1x2 . . . xn)1/Hn

And we haven√x1x2 . . . xn ≥

n∑nk=1 1/xk

= 1⇒ x1x2 . . . xn ≥ 1.

Combining the above two inequalities, we getn∑

k=1

1

k· xkk ≥ Hn.

• Let x, y ∈ R+ such that x+ y = 2. Prove that, x3y3(x3 + y3) ≤ 2.

• Prove that, n∏k=1

kk >(n+ 1

2

)n(n+1)2

• Suppose a, b, c are side lengths of a triangle. Prove that,(1 +

b− ca

)a(1 +

c− ab

)b(1 +

a− bc

)c

≤ 1.

6

1.4 Application of Cauchy-Schwarz (C-S)

All a, b, c, d in this subsection stand for positive real numbers, unlessotherwise specified.

(a2 + b2 + c2)(b2 + c2 + a2) ≥ (ab+ bc+ ca)2 ⇒ a2 + b2 + c2 ≥ ab+ bc+ ca.(bca

+ca

b+ab

c

)(cab

+ab

c+bc

a

)≥ (c+ a+ b)2 ⇒ bc

a+ca

b+ab

c≥ a+ b+ c.(

a2

b2+b2

c2+c2

a2

)(b2

c2+c2

a2+a2

b2

)≥(a

c+b

a+c

b

)2

⇒ a2

b2+b2

c2+c2

a2≥ a

c+b

a+c

b.

• Suppose x, y, z > 1 satisfy 1x

+ 1y

+ 1z

= 2. Prove that,

√x+ y + z ≥

√x− 1 +

√y − 1 +

√z − 1.

Solution: 1

x+

1

y+

1

z= 2⇒ x− 1

x+y − 1

y+z − 1

z= 1.

Then, x+ y + z =∑cyc

x ·∑cyc

(x− 1

x

) C−S≥ (

∑cyc

√x− 1)2.

• Let a, b, c be positive real numbers with a+ b+ c = 1. Prove that,

(a+ b)2(1 + 2c)(2a+ 3c)(2b+ 3c) ≥ 54abc.

• Let a1, . . . , an and b1, . . . , bn be real numbers and let A,B > 0 such thatA2 ≥ a21 + a22 + . . .+ a2n and B2 ≥ b21 + b22 + . . .+ b2n. Prove that,

(A2− a21− a22− . . .− a2n)(B2− b21− b22− . . .− b2n) ≤ (AB− a1b1− a2b2− . . .− anbn)2

Solution: Let a = a21 + . . .+ a2n and b = b21 + . . .+ b2n. If A2 = a or B2 = b thenthe given inequality holds trivially. So assume now A2 > a,B2 > b. Then,√

(A2 − a)(B2 − b) + a1b1 + . . .+ anbnC−S≤

√(A2 − a)(B2 − b) + ab

C−S≤

√(A2 − a+ a)(B2 − b+ b) = AB.

• Let a1, . . . , an and b1, . . . , bn be real numbers such that

(a21 + a22 + . . .+ a2n − 1)(b21 + b22 + . . .+ b2n − 1) > (a1b1 + a2b2 + . . .+ anbn − 1)2.

Prove that, a21 + a22 + . . .+ a2n > 1 and b21 + b22 + . . .+ b2n > 1.

• Let a, b, c be positive real numbers with abc = 1. Prove that,

a

a+ b4 + c4+

b

b+ c4 + a4+

c

c+ a4 + b4≤ 1.

Solution: (a+ b4 + c4)(a3 + 1 + 1)C−S≥ (a2 + b2 + c2)2 gives

∑cyc

a

a+ b4 + c4≤ a4 + b4 + c4 + 2(a+ b+ c)

(a2 + b2 + c2)2=a4 + b4 + c4 + 2abc(a+ b+ c)

(a2 + b2 + c2)2≤ 1,

Since a2b2 + b2c2 + c2a2 ≥ abc(a+ b+ c). (By C-S or AM-GM.)

• Let a, b, c ∈ R+ such that 1a+b+1

+ 1b+c+1

+ 1c+a+1

≥ 1. Prove that, a+ b+ c ≥ab+ bc+ ca.

7

• Let a, b, c be positive real numbers. Find the extreme values of the expression√a2x2 + b2y2 + c2z2 +

√b2x2 + c2y2 + a2z2 +

√c2x2 + a2y2 + b2z2

where x, y, z are real numbers such that x2 + y2 + z2 = 1.

Solution: Let us denote A = a2x2 + b2y2 + c2z2, B = b2x2 + c2y2 + a2z2 andC = c2x2+a2y2+b2z2. Note that A+B+C = (a2+b2+c2)(x2+y2+z2) = a2+b2+c2.

Now,√A+√B +

√C

C−S≤

√(1 + 1 + 1)(A+B + C) =

√3(a2 + b2 + c2).

And equality is achieved when x = y = z. So it is indeed the maximum value of thegiven expression. On the other hand, observe that,

AB = (a2x2 + b2y2 + c2z2)(b2x2 + c2y2 + a2z2)C−S≥ (abx2 + bcy2 + caz2)2 etc.

So that,√AB +√BC +

√CA ≥ (ab+ bc+ ca)(x2 + y2 + z2) = ab+ bc+ ca.

Hence,(√

A+√B +

√C)2

= A+B + C + 2(√

AB +√BC +

√CA

)≥ a2 + b2 + c2 + 2(ab+ bc+ ca) = (a+ b+ c)2.

Again, its easy to check that this minimum is achieved; for instance, whenx = 1, y = 0, z = 0.

1.4.1 Titu’s Lemma

This is a modification of C-S inequality, namely the following :

a21b1

+a22b2

+ . . .+a2nbn≥ (a1 + a2 + . . .+ an)2

b1 + b2 + . . .+ bn.

Where bk’s are positive real numbers and ak’s are any real numbers. Hereequality holds iff for some m, ak = mbk holds for each k. This form of C-S inequalityis more useful in certain type of problems, as seen below:

• Suppose a, b, c > −1 are real numbers such that a2 + b2 + c2 = 3. Prove that,1

1 + ab+

1

1 + bc+

1

1 + ca≥ 3

2.

Solution: 1

1 + ab+

1

1 + bc+

1

1 + ca

T itu≥ 9

3 +∑ab≥ 9

3 +∑a2

=3

2.

• For positive reals a, b, c with abc = 1, show that,

(a+ b− 1)2

c+

(b+ c− 1)2

a+

(c+ a− 1)2

b≥ a+ b+ c.

Solution: Its easy to get that a+ b+ c ≥ 3. Hence,∑cyc

(a+ b− 1)2

c

∑cyc

cT itu≥

[2(a+ b+ c)− 3

]2≥ (a+ b+ c)2.

• Let a, b, c be positive real numbers. Prove that,a

b+ 2c+

b

c+ 2a+

c

a+ 2b≥ 1.

Solution: ∑cyc

a

b+ 2c=∑cyc

a2

ab+ 2ca

T itu≥ (a+ b+ c)2

3(ab+ bc+ ca)≥ 1.

8

1.5 Some Special Topics

1.5.1 Difference Technique

This is a simple but amazing technique. In some situations, where direct AM-GM/C-S inequality does not work, but just a trick makes the solution magicallyfeasible.• Let a, b, c be positive real numbers with a+ b+ c = 3. Prove that,

a

1 + b2+

b

1 + c2+

c

1 + a2≥ 3

2.

Solution: a

1 + b2+

b

1 + c2+

c

1 + a2AM−GM≤ a

2b+

b

2c+

c

2a≥ 3

2!?

So simply applying AM-GM does not work here. However we can do like this:∑cyc

a

1 + b2=∑cyc

a− ab2

1 + b2≥∑cyc

a− ab2

2b= 3− ab+ bc+ ca

2.

And ab+ bc+ ca ≤ (a+b+c)2

3= 3. Hence L.H.S≥ 3− 3/2 = 3/2−Done!

So, just writing a1+b2

= a − ab2

1+b2, our solution is done−quite magically! This is

how this method works−we write each term as a difference of two terms and thenapply AM-GM/C-S.

• Suppose x, y, z > 1 satisfy 1x

+ 1y

+ 1z

= 2. Prove that,

√x+ y + z ≥

√x− 1 +

√y − 1 +

√z − 1.

Solution: Substitute√x− 1 = a etc. so that x = a2 + 1, y = b2 + 1, z = c2 + 1.

So we have to show√

3 + a2 + b2 + c2 ≥ a+ b+ c. Now,

2 =∑cyc

1

a2 + 1

T itu≥ 9

3 + a2 + b2 + c2⇒√

3 + a2 + b2 + c2 ≥ 3√2.

So we have to show 3√2≥ a + b + c. However, a2 + b2 + c2 ≥ 9

2− 3 = 3

2⇒

3(a2 + b2 + c2) ≥ 92

and 3(a2 + b2 + c2) ≥ (a+ b+ c)2 does not imply (a+ b+ c)2 ≤ 92.

Hence we proceed with the “diffrenence” approach :

2 =∑cyc

1

a2 + 1=∑cyc

(1− a2

a2 + 1

)⇒ 1 =

∑cyc

a2

a2 + 1≥ (a+ b+ c)2

3 + a2 + b2 + c2

Thus we get 3 + a2 + b2 + c2 ≥ (a + b + c)2 ⇒√

3 + a2 + b2 + c2 ≥ (a + b + c),as required !• Let a, b, c, d be positive real numbers with sum 4. Prove that,

a

1 + b2c+

b

1 + c2d+

c

1 + d2a+

d

1 + a2b≥ 2.

Solution: ∑cyc

a

1 + b2c=∑cyc

(a− ab2c

1 + b2c

)≥∑cyc

(a− ab2c

2b√c

)

= 4−∑cyc

b√a · ac2

≥ 4−∑cyc

b(a+ ac)

4= 4− 1

4

∑cyc

ab− 1

4

∑cyc

abc.

9

And, by AM-GM,∑cyc

ab ≤ 1

4

(∑cyc

a)2

= 4,∑cyc

abc ≤ 1

16

(∑cyc

a)3

= 4. Hence done.

• Let a, b, c ∈ R+ with a+ b+ c = 3. Prove that,

1

1 + 2b2c+

1

1 + 2c2a+

1

1 + 2a2b≥ 1.

• Let a, b, c, d ∈ R+ with a+ b+ c+ d = 4. Prove that,

a+ 1

b2 + 1+

b+ 1

c2 + 1+

c+ 1

d2 + 1+

d+ 1

a2 + 1≥ 4.

• Let a, b, c ∈ R+ with a2 + b2 + c2 = 3. Prove that,

1

a3 + 2+

1

b3 + 2+

1

c3 + 2≥ 1.

Solution: ∑cyc

1

a3 + 2=

1

2

∑cyc

(1− a3

a3 + 1 + 1

)≥ 1

2

∑cyc

(1− a3

3a

)= 1.

• Let a, b, c ∈ R+. Prove that,

2a√3a+ b

+2b√

3b+ c+

2c√3c+ a

≤√

3(a+ b+ c).

Solution:[∑cyc

a√3a+ b

]2 C−S≤

∑cyc

a ·∑cyc

a

3a+ b≤ 3

4

∑cyc

a ⇐⇒∑cyc

a

3a+ b≤ 3

4.

And,∑cyc

3a

3a+ b= 3−

∑cyc

b2

3ab+ b2T itu≤ 3− (a+ b+ c)2

3∑ab+

∑b2≤ 3− 3

4=

9

4.

1.5.2 An Useful Identity

◦ The identity(x+ y)(y + z)(z + x) = (x+ y + z)(xy + yz + zx)− xyz . . . (1)

is often useful, and along with 19(x+y+z)(xy+yz+zx)

AM−GM≥ xyz, it produces

another useful inequality

(x+ y)(y + z)(z + x) ≥ 8

9(x+ y + z)(xy + yz + zx) . . . (2)

• Let x, y, z be positive reals satisfying (x + y)(y + z)(z + x) = 1. Prove thatxy + yz + zx ≤ 3/4.

Solution:

x+ y + z =x+ y

2+y + z

2+z + x

2≥ 3

3

√(x+ y)(y + z)(z + x)

8=

3

2.

And, xyz ≤ (x+ y)(y + z)(z + x)

8=

1

8

So, 1 = (x+y+z)(xy+yz+zx)−xyz ≥ 32(xy+yz+zx)− 1

8⇒ xy+yz+zx ≤ 3

4.

• Suppose a, b, c are positive reals with abc = 1. Prove that,

10

√a+ b

a+ 1+

√b+ c

b+ 1+

√c+ a

c+ 1≥ 3.

• For a, b, c > 0, prove that,

3(a+ b)(b+ c)(c+ a) ≥ 8(a+ b+ c)(abc)2/3.

Solution:

(a+ b)(b+ c)(c+ a) ≥ 8

9(a+ b+ c)(ab+ bc+ ca)

AM−GM≥ 8

3(a+ b+ c)(abc)2/3.

• Let a, b, c be positive real numbers. Prove that,

a+ b+ c

3≤ 1

4

3

√(a+ b)2(b+ c)2(c+ a)2

abc.

Solution:(a+ b)2(b+ c)2(c+ a)2 ≥

(8

9(a+ b+ c)

)2

(ab+ bc+ ca)2

≥(

8

9(a+ b+ c)

)2

3abc(a+ b+ c)

= abc[4(a+ b+ c)

3

]3.

• If x, y, z be positive real numbers, prove that,

3(x2 + xy + y2)(y2 + yz + z2)(z2 + zx+ x2) ≥ (x+ y + z)2(xy + yz + zx)2

Solution: 4(x2 + xy + y2) = 3(x+ y)2 + (x− y)2 ≥ 3(x+ y)2. Hence,

LHS ≥ 34

43·[8

9(x+ y + z)(xy + yz + zx)

]2= (x+ y + z)2(xy + yz + zx)2.

• For a, b, c > 0, prove that,

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

]4/316

≥ abc(a+ b+ c).

• Let a, b, c be positive real numbers. Prove that,

3

√(a+ b)(b+ c)(c+ a)

8≥√ab+ bc+ ca

3.

◦ The inequality (x+ y)(y+ z)(z+x) ≥ 8xyz is very well-known, but the followingidentity is more precise:

(x+ y)(y + z)(z + x) = 8xyz + x(y − z)2 + y(z − x)2 + z(x− y)2

= x(y + z)2 + y(z + x)2 + z(x+ y)2 − 4xyz

◦ Often we have the condition/we assume1 that, xyz = 1. In that case, we havex+y+z ≥ 3, xy+yz+zx ≥ 3 and (x+y)(y+z)(z+x) = (x+y+z)(xy+yz+zx)−1.

1When the inequality is homogeneous in x, y, z, we may assume xyz = 1 without loss ofgenerality. See section *

11

• Let x, y, z be positive real numbers. Prove that,(x+

yz

x

)(y +

zx

y

)(z +

xy

z

)≥ 2(xyz + y2z + z2x+ x2y).

Solution: Dividing both sides by xyz the given inequality transforms into

(x2 + yz

zx

)(y2 + zx

xy

)(z2 + xy

yz

)≥ 2(1 +

y

x+z

y+x

z).

Now substitute a = x/y, b = y/z, c = z/x. Then the inequality is equivalent to

(a+ b)(b+ c)(c+ a) ≥ 2(1 + ab+ bc+ ca)

⇐⇒ (a+ b+ c)(ab+ bc+ ca)− 1 ≥ 2(1 + ab+ bc+ ca)

⇐⇒ (a+ b+ c− 2)(ab+ bc+ ca) ≥ 3.

This follows from a+ b+ c ≥ 3 and ab+ bc+ ca ≥ 3.

• Let a, b, c be positive real numbers. Prove that,√(a2b+ b2c+ c2a)(ab2 + bc2 + ca2) ≥ abc+ 3

√(a3 + abc)(b3 + abc)(c3 + abc).

Solution: Here also we divide both sides by abc and substitute x = a/b, y =b/c, z = c/a. Then xyz = 1 and hence the inequality transforms into√

(x+ y + z)(xy + yz + zx) ≥ 1 + 3

√(x/z + 1)(y/x+ 1)(z/y + 1)

⇐⇒√

(x+ y)(y + z)(z + x) + 1 ≥ 1 + 3

√(x+ y)(y + z)(z + x)

So if 3

√(x+ y)(y + z)(z + x) = t then we need to show

√t3 + 1 ≥ 1 + t.

Now, t3 = (x+ y)(y + z)(z + x) ≥ 8xyz = 8⇒ t ≥ 2.

Hence,√t3 + 1 =

√(t+ 1)(t2 − t+ 1) ≥

√(t+ 1)(2t− t+ 1) = t+ 1.

• Let a, b, c be positive real numbers with abc = 1. Prove that,

(a+ b)(b+ c)(c+ a) ≥ 4(a+ b+ c− 1).

Solution: The given inequality is equivalent to

(a+ b+ c)(ab+ bc+ ca)− 1 ≥ 4(a+ b+ c− 1)

or, ab+ bc+ ca+3

a+ b+ c≥ 4.

Now,3ab+ bc+ ca

3+

3

a+ b+ c≥ 4

[(ab+ bc+ ca)3

9(a+ b+ c)

]1/4.

And (ab+bc+c)2 ≥ 3abc(a+b+c) = a+b+c and (ab+bc+ca) ≥ 3(abc)2/3 = 3.Multiplying these two, we get (ab+ bc+ ca)3 ≥ 9(a+ b+ c). Hence done.

12

1.5.3 The Substitution Strategy

1.5.4 Rearrangement Inequality

The rearrangement inequality is a remarkable inequality, although it is hardly used.Given two sequences of real numbers, namely a1, a2, . . . an and b1, b2, . . . , bn, it saysthat the sum a1b1 + a2b2 + . . . + anbn will be maximum when the sequences aresimilarly ordered and minimum when the sequences are oppositely ordered. Thisinequality can be used to prove Nesbitt’s inequality, Tchebyshev’s inequality, C-S inequality (try it). The following examples illustrate use of this inequality inolympiad problems.• Let x1, x2, . . . , xn be distinct positive integers. Find the minimum value of :

x112

+x222

+ . . .+xnn2.

Solution: Let (y1, y2, . . . , yn) be the permutation of (x1, x2, . . . , xn) such that y1 <y2 < . . . < yn. Clearly, y1 ≥ 1 . y2 > y1 ⇒ y2 ≥ 2. Inductively one can get yk ≥ k forall k = 1, 2, . . . , n. Now, note that the sequences (y1, y2, . . . , yn) and (1, 1

22, . . . , 1

n2 )are oppositely ordered. Hence, by Rearrangement inequality,

x112

+x222

+ . . .+xnn2≥ y1

12+y222

+ . . .+ynn2

≥ 1

12+

2

22+ . . .+

n

n2= 1 +

1

2+ . . .+

1

n.

Equality occurs when xk = k for each k = 1, 2, . . . , n. [Since equality occurs it isindeed the minimum value of the given expression.]• Suppose a, b, c are side-lengths of a triangle. Prove that,

a2(b+ c− a) + b2(c+ a− b) + c2(a+ b− c) ≤ 3abc.

Solution: First note that,

a(b+ c− a) ≤ b(c+ a− b) ⇐⇒ 0 ≤ (a− b)(a+ b− c).Thus a(b+ c− a), b(c+ a− b), c(a+ b− c) and a, b, c are ordered oppositely. Hence,∑

cyc

a2(b+ c− a) ≤∑cyc

b · a(b+ c− a)

∑cyc

a2(b+ c− a) ≤∑cyc

c · a(b+ c− a)

Just adding these two inequalities and collecting terms on the RHS,

2∑cyc

a2(b+ c− a) ≤∑cyc

ba(b+ c− a+ c+ a− b) = 6abc. Done!

• Let a, b, c be side-lengths of a triangle. Prove that,

a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.

Solution: Following is possibly the simplest proof of this famous problem.∑cyc

1

a· a(b+ c− a) ≥

∑cyc

1

c· a(b+ c− a)

⇒ a+ b+ c ≥ a+ b+ c+a(b− a)

c+b(c− b)

a+c(a− c)

b⇒ a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.

13

1.6 Manipulations

Here we shall see some inequality problems which are solved by clever manipulations.Sometimes substitutions make the manipulations much easier and shorter.• For a, b, c, x, y, z ≥ 0, prove that,

(a2 + x2)(b2 + y2)(c2 + z2) ≥ (ayz + bzx+ cxy − xyz)2.

Solution: Note that the given inequality holds trivially when one or more amongx, y, z is zero. Now assume xyz 6= 0. Let u = a/x, v = b/y, w = c/z. Then theinequality is equivalent to

(u2 + 1)(v2 + 1)(w2 + 1) ≥ (u+ v + w − 1)2.

⇐⇒ u2v2w2 + u2v2 + v2w2 + w2u2 + 2(u+ v + w) ≥ 2(uv + vw + wu).

which is true since AM-GM gives u2v2 + u+ v ≥ 3uv ≥ 2uv etc.

• For a, b, c > 0 with abc = 1, prove that,

3√a+

3√b+ 3√c ≤ 3

√3(3 + a+ b+ c+ ab+ bc+ ca).

Solution: Cubing both sides, the inequality is equivalent to (using abc = 1)

3(3√a2b+

3√ab2 +

3√b2c+

3√bc2 +

3√a2c+

3√ac2) ≤ 3 + 3(ab+ bc+ ca) + 2(a+ b+ c).

To prove it, first note that ab+ bc+ ca ≥ 3(abc)2/3 = 3. Hence,

3 + 3(ab+ bc+ ca) + 2(a+ b+ c) ≥ 6 + 2(ab+ bc+ ca) + 2(a+ b+ c).

=∑cyc

(a+ ab+ 1) +∑cyc

(a+ ac+ 1) ≥ 3∑cyc

3√a2b+ 3

∑cyc

3√a2c, done!

• For a, b, c > 0, prove that,

a2

b2+b2

c2+c2

a2≥ a

b+b

c+c

a.

Solution: Since the given inequality is cyclic in a, b, c (i.e. it remains unchangedwhen the role of any two of a, b, c is intechanged.) so we can assume without lossof generality that a ≥ b ≥ c. Now,

a2

b2+b2

c2+c2

a2−(a

b+b

c+c

a

)=

a(a− b)b2

+b(b− c)c2

+c(c− a)

a2

=a(a− b)

b2+b(b− c)c2

+c(c− b+ b− a)

a2

= (a− b)(a

b2− c

a2

)+ (b− c)

(b

c2− c

a2

)=

(a− b)a2b2

(a3 − b2c) +(b− c)c2a2

(a2b− c3)

Now the ordering a ≥ b ≥ c gives a3 ≥ b2c and a2b ≥ c3. Hence the expressionabove is non-negative! Note that the key-step was writing c − a as c − b + b − a.This idea is often useful, while proving cyclic inequalities by brute-force.

14

• If a, b, c be sides of a triangle, then show that,

a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.

Solution 1: Here also we can assume w.l.o.g. that a ≥ b ≥ c.

a2b(a− b) + b2c(b− c) + c2a(c− a) = a2b(a− b) + b2c(b− c) + c2a(c− b+ b− a)

= (a2b− c2a)(a− b) + (b2c− c2a)(b− c)= a(ab− c2)(a− b) + c(b2 − ca)(b− c)

Now, if a−b ≤ b−c, then we get 2b ≥ c+a⇒ b2 ≥(a+c2

)2≥ ca. And a ≥ b ≥ c

gives ab ≥ c2. Thus the expression above, is non-negative.Else, a − b < b − c. In that case, a(ab − c2)(a − b) + c(b2 − ca)(b − c) ≥

c(b− c)(ab− c2 + b2 − ca) = c(b− c)2(a+ b+ c) ≥ 0.Solution 2: (official solution, imo 1983) For any triangle of sides a, b, c there

exists 3 non-negative numbers x, y, z such that a = y+ z, b = z+ x, c = x+ y(thesenumbers correspond to the division of the sides of the triangle by the points ofcontact of the incircle.) The inequality becomes

(y + z)2(z + x)(y − x) + (z + x)2(x+ y)(z − y) + (x+ y)2(y + z)(x− z) ≥ 0.

Expanding, we get xy3 +yz3 +zx3 ≥ xyz(x+y+z). This follows from Cauchy’sinequality (xy3 + yz3 + zx3)(z + x + y) ≥ (

√xyz(y + z + x))2 with equality if and

only if xy3/z = yz3/x = zx3/y, or equivalently x = y = z i.e. a = b = c.

Note: This a = y + z, b = z + x, c = x + y substitution is named as ‘Ravi’s Substitution.’

I would like to share how one can arrive at such ‘equivalent’ forms. Here is mymanipulations for this problem:

∑cyc

a2b(a− b) ≥ 0Ravi′s Subst.←→

∑cyc

(y + z)2(z + x)(y − x) ≥ 0

↔∑cyc

(y + z)2(yz + xy) ≥∑cyc

(y + z)2(zx+ x2)

↔∑cyc

(y3z + 2y2z2 + yz3 + xy3 + 2xy2z + xyz2)

≥∑cyc

(y2zx+ 2yz2x+ z3x+ y2x2 + 2x2yz + z2x2)

↔∑cyc

y3z + 2∑cyc

yz3 + 2∑cyc

x2y2 + 3∑cyc

xy2z ≥ 5∑cyc

xy2z + 2∑cyc

x2y2 +∑cyc

y3z

↔ 2∑cyc

yz3 ≥ 2∑cyc

x2yz ↔ x2

y+y2

z+z2

x≥ x+ y + z.

Solution 3: Assume a ≥ b ≥ c. Now, the given inequality is equivalent to,

a(b− c)2(b+ c− a) + b(a− b)(a− c)(a+ b− c) ≥ 0.

15

which is certainly true. This Solution is due to a contestant of imo’83 and he wasawarded a medal for this amazing identity! I don’t know what was the motivationbehind it, anyways, here is my manipulations to arrive at that:

a2b(a− b) + b2c(b− c) + c2a(c− a)

= a2b(a− b)− abc2 − ab2c+ b2c(a+ b− c) + c2a(b+ c− a)

= ab(a2 − ab− c2 − bc) + b2c(a+ b− c) + c2a(b+ c− a)

= ab(a+ c)(a− c− b) + b2c(a+ b− c) + c2a(b+ c− a)

= b2c(a+ b− c) + a(b+ c− a)(c2 − ba− bc)= b2c(a+ b− c) + a(b+ c− a)(−b(a+ b− c) + c2 − 2bc+ b2)

= b(a+ b− c)(bc− a(b+ c− a)) + a(b+ c− a)(c2 − 2bc+ b2)

= b(a+ b− c)(a− b)(a− c) + a(b+ c− a)(b− c)2.

Solution 4: With the same approach, but simpler calculations:

a2b(a− b) + b2c(b− c) + c2a(c− a)

= a2b(a− b) + b2c(b− c) + c2a(c− a)− ab2[(a− b) + (b− c) + (c− a)]

= (a2b− ab2)(a− b) + (b2c− ab2)(b− c) + (c2a− ab2)(c− a)

= ab(a− b)2 + (b− c)[b(bc− ab)− a(c− a)(b+ c)]

= ab(a− b)2 + (b− c)(a− c)(ab+ ac− b2)≥ ab(a− b)2 + (b− c)(a− c)(b+ c− a)a ≥ 0. (assuming a ≥ b ≥ c.)

• For any three real numbers a, b, c, prove that,

(a2 + b2 + c2)2 ≥ 3(a3b+ b3c+ c3a).

Solution 1: Use (x+ y + z)2 ≥ 3(xy + yz + zx) for

x = a2 + bc− ab, y = b2 + ca− bc, z = c2 + ab− ca.

Solution 2: (Due to Vasc)∑cyc

a4 + 2∑cyc

a2b2 − 3∑cyc

a3b =1

2

∑cyc

(a2 − 2ab+ bc− c2 + ca)2 ≥ 0

• For a, b, c ∈ R+, prove that,

a+ b

b+ c+b+ c

c+ a+c+ a

a+ b+

3(ab+ bc+ ca)

(a+ b+ c)2≥ 4.

• If a, b, c be sides of a triangle, then show that,

(b+ c)2

a2 + bc+

(c+ a)2

b2 + ca+

(a+ b)2

c2 + ab≥ 6.

• For a, b, c ∈ R+, prove that,

a

b+b

c+c

a≥ a+ b

b+ c+b+ c

c+ a+c+ a

a+ b.

• Let a, b, c ∈ R+ such that a+ b+ c+ 1 = 4abc. Prove that,

1

a4 + b+ c+

1

b4 + c+ a+

1

c4 + a+ b≤ 3

a+ b+ c.

16

2 Problems

Tactics are best learnt through problems. Here are a handful of problems collectedfrom various contests and books.• If a, b, c ∈ R such that a+ b+ c ≥ abc, prove that,

a2 + b2 + c2 ≥√

3abc.

• For any real numbers x, y, z prove that,

x4 + y4 + z2 + 1 ≥ 2x(xy2 − x+ z + 1).

• Let a, b, c, d ∈ R+ such that abcd = 1. Prove that,

1

a+ b+ 2+

1

b+ c+ 2+

1

c+ d+ 2+

1

d+ a+ 2≤ 1.

• a, b, c ∈ R+ satisfy abc(a+ b+ c) = 3. Prove that,

(b+ c)(c+ a)(a+ b) ≥ 8.

• For a, b, c ∈ R+, Prove that,√a

b+ c+

√b

c+ a+

√c

a+ b> 2.

• For a, b, c ∈ R+ such that ab+ bc+ ca ≤ 3abc, prove that,

a+ b+ c ≤ a3 + b3 + c3.

• For positive reals a, b, c such that abc = 1, prove that,

ab

a5 + b5 + ab+

bc

b5 + c5 + bc+

ca

c5 + a5 + ca≤ 1.

• Suppose a, b, c > 1 satisfy

1

a2 − 1+

1

b2 − 1+

1

c2 − 1= 1.

Then show that,1

a+ 1+

1

b+ 1+

1

c+ 1≤ 1.

• Prove that in any non-obtuse triangle ABC,

sinA+ sinB + sinC > 2.

• If a, b, c > 0, prove that

(aabbcc)1

a+b+c ≥ 3√abc.

17

• Let a, b, c, d be positive reals such that, a+ b+ c+ d = 4. Prove that,

1

a2 + 1+

1

b2 + 1+

1

c2 + 1+

1

d2 + 1≥ 2.

• Let a, b, c be sides of a triangle corresponding to angles α, β, γ respectively.Prove that,

π

3≤ aα + bβ + cγ

a+ b+ c<π

2.

• For a, b, c, p, q > 0 with abc = 1 and p > q, prove that,

p(a2 + b2 + c2) + q(1

a+

1

b+

1

c

)≥ (p+ q)(a+ b+ c).

• For a, b, c > 0, prove that,

a+ b+ c3√abc

+8abc

(a+ b)(b+ c)(c+ a)≥ 4.

• Suppose x, y, z ∈ R such that x+ y + z = 0. Prove that,

x(x+ 2)

2x2 + 1+y(y + 2)

2y2 + 1+z(z + 2)

2z2 + 1≥ 0.

• Let a, b, c ∈ R+. Prove that,

1√a3 + b

+1√b3 + c

+1√c3 + a

≤ 1√2

(1

a+

1

b+

1

c

).

• For 0 < x < 1 prove that,

x+1

xx< 2.

• Let a, b be positive reals. Prove that,

a2

a+ b+ 1+

b+ 1

a2 + a> 1.

• For n ≥ 3, auppose a2, a3, . . . , an ∈ R+ such that a2a3 . . . an = 1. Prove that,

(1 + a2)2(1 + a3)

3 . . . (1 + an)n > nn.

• Suppose a, b, c, d ∈ R+ such that abcd = 1 and

a+ b+ c+ d >a

b+b

c+c

d+d

a. Prove that, a+ b+ c+ d <

b

a+c

b+d

c+a

d.

• Let a, b, c be positive real numbers. Prove that,

1

a+ 1b

+ 1+

1

b+ 1c

+ 1+

1

c+ 1a

+ 1≥ 3

3√abc+ 1

3√abc

+ 1.

18

3 Solutions

Most of the solution here are shortened, in order to save space. In competitions youneed to write more elaborately1.• If a, b, c ∈ R such that a+ b+ c ≥ abc, prove that,

a2 + b2 + c2 ≥√

3abc.

Solution: Using x2 + y2 + z2 ≥ xy + yz + zx twice, we get

(a2 + b2 + c2)2 ≥ 3(a2b2 + b2c2 + c2a2) ≥ 3abc(a+ b+ c) ≥ 3(abc)2.

• For any real numbers x, y, z prove that,

x4 + y4 + z2 + 1 ≥ 2x(xy2 − x+ z + 1).

Solution: The given inequality is equivalent to

(x2 − y2)2 + (x− z)2 + (x− 1)2 ≥ 0.

• Let a, b, c, d ∈ R+ such that abcd = 1. Prove that,

1

a+ b+ 2+

1

b+ c+ 2+

1

c+ d+ 2+

1

d+ a+ 2≤ 1.

Solution:

1

a+ b+ 2+

1

c+ d+ 2≤ 1

2(√ab+ 1)

+1

2(√cd+ 1)

=1

2(√ab+ 1)

+

√ab

2(√ab+ 1)

=1

2.

Here we used√cd = 1√

ab. In a similar way, we can prove

1

b+ c+ 2+

1

d+ a+ 2≤ 1

2.

• a, b, c ∈ R+ satisfy abc(a+ b+ c) = 3. Prove that,

(b+ c)(c+ a)(a+ b) ≥ 8.

Solution:3 = abc(a+ b+ c) ≥ abc · 3 3

√abc ⇒ 1 ≥ abc.

(ab+ bc+ ca)2 ≥ 3(ab · bc+ bc · ca+ ca · ab) = 3abc(a+ b+ c) = 9.

Hence, ab+ bc+ ca ≥ 3⇒ (a+ b+ c) ≥√

3(ab+ bc+ ca ≥√

3 · 3 = 3.

So, (b+ c)(c+ a)(a+ b) = (a+ b+ c)(ab+ bc+ ca)− abc ≥ 3 · 3− 1 = 8.• For a, b, c ∈ R+, Prove that,

1You can view the official solutions to the past papers of RMO-INMO (or IMO), to get an ideaabout how to write proofs in a systematic way.

19

√a

b+ c+

√b

c+ a+

√c

a+ b> 2.

Solution: We use 1 + x ≥ 2√x for x = b+c

ato obtain

a+ b+ c

a≥ 2

√b+ c

a⇒∑cyc

√a

b+ c≥∑cyc

2a

a+ b+ c= 2.

For equality to hold, we must have b+ca

= c+ab

= a+bc

= 1 which leads toa+ b+ c = 0, not possible for a, b, c ∈ R+.• For a, b, c ∈ R+ such that ab+ bc+ ca ≤ 3abc, prove that,

a+ b+ c ≤ a3 + b3 + c3.

Solution: ab+ bc+ ca ≤ 3abc⇒ abc(a+ b+ c) ≤ (ab+ bc+ ca)2 ≤ (3abc)2.

⇒ a+ b+ c ≤ 3abc ≤ a3 + b3 + c3.

• For positive reals a, b, c such that abc = 1, prove that,

ab

a5 + b5 + ab+

bc

b5 + c5 + bc+

ca

c5 + a5 + ca≤ 1.

Solution: (a3 − b3)(a2 − b2) ≥ 0⇒ a5 + b5 ≥ a2b2(a+ b). Hence

∑cyc

ab

a5 + b5 + ab≤∑cyc

ab

a2b2(a+ b) + ab=∑cyc

c

a+ b+ c= 1.

• Suppose a, b, c > 1 satisfy

1

a2 − 1+

1

b2 − 1+

1

c2 − 1= 1.

Then show that,1

a+ 1+

1

b+ 1+

1

c+ 1≤ 1.

Solution:

Subtitute1

a2 − 1= x,

1

b2 − 1= y,

1

c2 − 1= z, so that a =

√x+ 1

xetc.

Hence the given condition converts into x+ y + z = 1 and we have∑cyc

1

a+ 1=∑cyc

√x(√x+ 1−

√x)

AM−GM≤

∑cyc

(√x+ 1

2

)2

=x+ y + z + 3

4= 1.

Equality holds when x = y = z = 1/3 i.e. a = b = c = 2.• Prove that in any non-obtuse triangle ABC,

sinA+ sinB + sinC > 2.

20

Solution:sinA+ sinB + sinC > sin2A+ sin2B + sin2C = 2 + 2 cosA cosB cosC ≥ 2.

• If a, b, c > 0, prove that,

(aabbcc)1

a+b+c ≥ 3√abc.

Solution: Suppose (aabbcc)1

a+b+c < 3√abc. Then,

3 = a1

a+ b

1

b+ c

1

c≥ (a+ b+ c)

(1

aabbcc

) 1a+b+c

> (a+ b+ c)1

3√abc≥ 3

Thus we reached at 3 > 3, a contradiction!• Let a, b, c, d be positive reals such that, a+ b+ c+ d = 4. Prove that,

1

a2 + 1+

1

b2 + 1+

1

c2 + 1+

1

d2 + 1≥ 2.

Solution:

1− 1

1 + a2=

a2

1 + a2≤ a2

2a=a

2⇒∑ 1

1 + a2≥ 4− a+ b+ c+ d

2= 2.

• Let a, b, c be sides of a triangle corresponding to angles α, β, γ respectively.Prove that,

π

3≤ aα + bβ + cγ

a+ b+ c<π

2.

Solution: Since a, b, c and α, β, γ are similarly ordered, so Tchebyshev’s inequal-ity gives

α + β + γ

3· a+ b+ c

3≤ aα + bβ + cγ

3⇒ π

3≤ aα + bβ + cγ

a+ b+ c.

Next, a < b+ c⇒ 2a < a+ b+ c etc. Therefore,

2(aα + bβ + cγ

a+ b+ c

)=

2a

a+ b+ c· α +

2b

a+ b+ c· β +

2c

a+ b+ c· γ < α + β + γ = π.

• For a, b, c, p, q > 0 with abc = 1 and p > q, prove that,

p(a2 + b2 + c2) + q(1

a+

1

b+

1

c

)≥ (p+ q)(a+ b+ c).

Solution 1: First note that abc = 1 gives a+ b+ c ≥ 3 and ab+ bc+ ca ≥ 3.

Now,∑cyc

(a− 1)2 ≥ 0⇒∑cyc

a2 −∑cyc

a ≥∑cyc

a− 3 ≥∑cyc

a−∑cyc

ab.

So, q(∑

cyc

a−∑cyc

ab)≤ q

(∑cyc

a2 −∑cyc

a)≤ p

(∑cyc

a2 −∑cyc

a)

(∗)

But, last inequality is true only if∑

cyc a2 ≥ ∑cyc a. However, we can prove it :

21

a2 + b2 + c2 ≥ a+ b+ c

3(a+ b+ c) ≥ a+ b+ c

And using abc = 1 we can rewrite (∗) as p∑a2 + q

∑ 1a≥ (p+ q)

∑a.

Solution 2: By AM-GM and using abc = 1,∑cyc

a2 + a2 + bc ≥∑cyc

3a and∑cyc

bc+ ca+ c2 ≥∑cyc

3c

Adding them, we get a2 + b2 + c2 + ab + bc + ca ≥ 2(a + b + c). Now we canproceed as in solution 1.• For a, b, c > 0, prove that,

a+ b+ c3√abc

+8abc

(a+ b)(b+ c)(c+ a)≥ 4.

Solution: By weighted AM-GM (with 3 and 1 as weights)

3a+ b+ c

3 3√abc

+8abc

(a+ b)(b+ c)(c+ a)≥ 4

(8(a+ b+ c)3

27(a+ b)(b+ c)(c+ a)

)4

.

And,

8(a+ b+ c)3 = [(b+ c) + (c+ a) + (a+ b)]3AM−GM≥ 27(a+ b)(b+ c)(c+ a).

• Suppose x, y, z ∈ R such that x+ y + z = 0. Prove that,

x(x+ 2)

2x2 + 1+y(y + 2)

2y2 + 1+z(z + 2)

2z2 + 1≥ 0.

Solution: ∑cyc

x(x+ 2)

2x2 + 1≥ 0 ⇐⇒

∑cyc

2x(x+ 2)

2x2 + 1+ 1 ≥ 3

⇐⇒∑cyc

(2x+ 1)2

2x2 + 1≥ 3.

Now, x2 = (y + z)2 ≤ 2(y2 + z2)⇒ 2x2 ≤ 43(x2 + y2 + z2). Hence,

∑cyc

(2x+ 1)2

2x2 + 1≥∑cyc

3(2x+ 1)2

4(x2 + y2 + z2) + 3= 3.(as x+ y + z = 0.)

• Let a, b, c ∈ R+. Prove that,

1√a3 + b

+1√b3 + c

+1√c3 + a

≤ 1√2

(1

a+

1

b+

1

c

)Solution:∑

cyc

1√a3 + b

AM−GM≤

∑cyc

1√2

1

a3/4b1/4

Weighted AM−GM

≤ 1√2

∑cyc

1

4

(3

a+

1

b

).

• For 0 < x < 1 prove that,

x+1

xx< 2.

22

Solution: By weighted AM-GM, (Using 1− x and x as weights!) we have

11−x(

1

x

)x

≤ (1− x) · 1 + x · 1

x⇒ 1

xx≤ 2− x.

• Let a, b be positive reals. Prove that,

a2

a + b + 1+

b + 1

a2 + a> 1

Solution:

a2

a+ b+ 1+

b+ 1

a2 + a=

a2 + a

a+ b+ 1+a+ b+ 1

a2 + a− 2 +

b+ 1

a+ b+ 1+

a

a+ 1

≥ b+ 1

a+ b+ 1+

a

a+ 1>

b+ 1

a+ b+ 1+

a

a+ b+ 1= 1.

• For n ≥ 3, auppose a2, a3, . . . , an ∈ R+ such that a2a3 . . . an = 1. Prove that,

(1 + a2)2(1 + a3)

3 . . . (1 + an)n > nn.

Solution: By weighted AM-GM,

(k − 1)1

k − 1+ ak ≥ k

(ak

(k − 1)k−1

)k

⇒ (1 + ak)k ≥ kk

(k − 1)k−1· ak

So,n∏

k=2

(1 + ak)k ≥n∏

k=2

kk

(k − 1)k−1· ak = nn · a2a3 . . . an = nn

Easy to see that equality does not hold here.• Suppose a, b, c, d ∈ R+ such that abcd = 1 and

a + b + c + d >a

b+

b

c+

c

d+

d

a. Prove that, a + b + c + d <

b

a+

c

b+

d

c+

a

d.

Solution:

3(a + b + c + d) +

(b

a+

c

b+

d

c+

a

d

)> 3

(a

b+

b

c+

c

d+

d

a

)+

(b

a+

c

b+

d

c+

a

d

)

=∑cyc

(a

b+

a

b+

b

c+

a

d

)≥∑cyc

44

√a3

bcd= 4(a + b + c + d). (using abcd = 1)

• Let a, b, c be positive real numbers. Prove that,

1

a+ 1b

+ 1+

1

b+ 1c

+ 1+

1

c+ 1a

+ 1≥ 3

3√abc+ 1

3√abc

+ 1.

Solution: We have the identity∑cyc

b

ab+ b+ 1= 1− (abc− 1)2

(ab+ b+ 1)(bc+ c+ 1)(ca+ a+ 1)

And, by Holder′s inequality,

(ab+ b+ 1)(bc+ c+ 1)(ca+ a+ 1) ≥ [(ab · bc · ca)1/3 + (b · c · a)1/3 + 1]3

So, if 3√abc = m then combining the above two, we get,∑

cyc

b

ab+ b+ 1≥ 1− (m3 − 1)2

(m2 +m+ 1)3= 1− (m− 1)2

m2 +m+ 1=

3m

m2 +m+ 1.

23