My Inequality Project (1)

7/28/2019 My Inequality Project (1)

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My Inequality Project

M.Ramchandran

No.33,Belfast Apts,Ramachandran Street,T.Nagar

E-mail address: [email protected]: http://mathematicaldreams.wordpress.com/

Dedicated to Professor XY


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Contents

Preface v

Chapter 1. AM-GM Inequality 11. Arithmetic Mean - Geometric Mean 12. Proof 23. Beginners Practice Problems 44. Geometric Interpretations 75. AM-GM Tautogrid Technique 76. Nessbits Inequality 9

7. The Reverse Technique 108. The Weighted AM-GM Inequality 159. Method Of Balancing Co-efficients by AM-GM 1610. Quasiliearisation 1811. Equivalent Summation Technique 1912. The G function 2113. Problem Set 22

iii


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Preface

This is the preface and it is created using a TeX field in a paragraph by itself containing \chapter*{Preface}.When the document is loaded, this appears if it were a normal chapter, but it is actually an unnumberedchapter.

v


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CHAPTER 1

AM-GM Inequality

1. Arithmetic Mean - Geometric Mean

Lets go from scratch,

Definition 1 (Arithmetic Mean). Arithmetic mean of two non-negative real numbers a and b isdefined as the average of the two numbers and is mathematically expressed as -

A.M. =a + b

2

where ofcourse,

A.M. stands for the arithmetic mean of the two concerned non-negative real numbers - a and b.

Extending this idea to n-variables - a1, a2, a3, . . ,an we get that -

A.M. =a1 + a2 + a3 + ... + an

n=

ni=1 ai

n

Definition 2 (Geometric Mean). Geometric Mean of two real numbers is the collection of positivereal numbers is the nth root of the product of the numbers. Note that if it is even, we take the positive nth

root and it is mathematically expressed as -

G.M. =

ab

where,G.M. stands for Geometric Mean of the two concerned non-negative real numbers - a and b

Extending this idea to n-variables - a1, a2, a3, . . ,an we get that -

G.M. = (a1a2a3...an)1n = (

ni=1

ai)1n

Recall the fact that for any real number x , we know that x2 0.therfore we know that for all non-negative real numbers -

a and

b -

(

a

b)2 0a + b 2

ab 0

a + b

2

ab

ring any bells?yes you have probably spotted that LHS of the inequality is the arithmetic mean we discussed and

ofcourse the RHS is the geometric mean. Now this encourages the following proposition -

Theorem 1. Arithmetic Mean of some n non-negative real numbers is always greatern than or equalto the Geometric Mean of the same

1


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2 1. AM-GM INEQUALITY

is that true? if so where is the validity? by seeing that it is true for certaina two positive reals doesntimply a wider truth for any number of variables.

2. Proof

Proof.

Lets proceede by proving the above statement for smaller numbers and eventually the generalcase-Since we know that A.M. G.M. for two variables we have -

p + q

2 pq

r + s

2 rs

also we have - pq+

rs

2

pqrs = (pqrs)14

combining we get that-

p + q+ r + s

4=

p+q2

+ r+s2

2

pq+

rs

2 (pqrs) 14

or,..p + q+ r + s

4 (pqrs) 14

which is AM-GM inequality for 4 variables!!!With a similar idea we can do the same as above for 8 - variables by splitting into 4-4 and using AM-GM

for 4 variables.Now something should be irking in your mind.... Cant we prove in the same way for any n?? - the

answer is NO. Why not? - well this covers only 2-powers not even even integers.. so this proof is incompleteconsidering the general case. But a proof with the same idea can be given by induction for any 2-powers.Itgoes as follows -

Consider 2k+1 variables - a1, a2, a3,...,a2k+1Assume the truth of the statement for 2k, we shall prove it for 2k+1

Since it is true for 2k, we have -

a1 + a2 + a3 + ... + a2k

2k (a1a2a3..a2k)

1

2k (1)and,

a2k+1 + a2k+2 + a2k+3 + ... + a2k+1

2k (a2k+1a2k+2a2k+3...a2k+1)

1

2k (2)again,

(1) + (2)

2=

a1 + a2 + ... + a2k+1

2k+1 (a1a2a3..a2k)

1

2k + (a2k+1a2k+2a2k+3...a2k+1)1

2k

2 (a1a2...a2k+1)

1

2k+1

or,a1 + a2 + a3 + ... + a2k+1

2k+1 (a1a2a3..a2k+1)

1

2k+1

thus by induction it is proved for all powers of 2!!! this is a bit closer to the full proof..

But then what about 6, 10.. and 3, 5, 7... how do we prove them..?with a little bit of trickery we get away with 3 - since the inequality is true for 4 we have -

a + b + c + (abc)13

4 (abc) 43 14 = (abc) 13

and this re-arranges to -a + b + c

3 (abc) 13


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2. PROOF 3

which is the AM-GM inequality for 3 reals!!!! this can also be done by the following way -

a + b + c + ( a+b+c3

)

4 [abc ( a + b + c

3]14

that is..a + b + c

3 [abc

(

a + b + c

3

]14

which re-arranges to -a + b + c

3 (abc) 13

The above two proofs crucially modifies our way of looking at AM- GM inequality. Let us re-defineourselves the meaning of AM,GM into -

A.M. = sum equaliserG.M. = product equaliserwhat does that mean ? It means that say A is the A.M. of 3 variables x, y, z and G the G.M. then -

x + y + z = 3A

and,

x.y.z = G3

If we know that the inequality is true for a certain m , then for any n such that m > n we can provethe validity of AM-GM inequality as follows-

a1 + a2 + ... + an + (m n)Am

(a1a2a3..an) 1m Amn

m

mA

m (G) nm Amnm

Am Gn Amnor,

An Gn= A G

thus we have prove the AM-GM inequality for any n as we know it to be true for any 2k.. or -

A.M. G.M.for any n This is trivial looking inequality is probably the most celebrated of all that we all shall discuss inthis book.It has far and wide applications..

Proof. The inequality is true for 2n if it is true for n or it is true for all powers of 2 (already proved)Suppose that the inequality is true for n numbers.We then choose

an =s

n 1where,

s = a1 + a2 + a3 + .... + an

According to the inductive hypothesis, we get

s +s

n 1 n(

a1a2a2a3...an sn 1

)1n

s (n 1)(a1a2a3..an) 1n1Therfore if the inequality is true for n numbers than it will be true for n 1 numbers and by induction(Cauchy Induction),the inequality is true for every natural number n .Equality occurs if and only if a1 =a2 = a3 = ... = an


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3. Beginners Practice Problems

1. For a,b,c R+0 .Prove that(a + b)(b + c)(c + a) 8abc

Solution

Note that by AM-GM,a + b 2

ab

b + c 2

bc

c + a 2ca

multiplying the above inequalities we get the desired with equality for a = b = c

2.For a,b,c R+0 . Prove thata2 + b2 + c2 ab + bc + ca

Solution

a2 + b2 2abb2 + c2 2bcc2 + a2 2ac

adding this we get the desired with equality for a = b = c

3. For a,b,c R+ such that - a + b + c = 2.Prove that

abc 8(1 a)(1 b)(1 c)Solution

Set 1 a = x, 1 b = y, 1 c = ztherfore by condition,

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

substituting this in the inequality we get that it is equivalent to -

(x + y)(y + z)(z + x) 8xyz

which we just proved! with equality for a = b = c = 23

4. For a,b,c R+.Prove that -

(a2b + b2c + c2a)(a2c + b2a + c2b) 9a2b2c2

Solution Observe that by AM-GM-

a2b + b2c + c2a 3abc


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3. BEGINNERS PRACTICE PROBLEMS 5

and,a2c + b2a + c2b 3abc

multiply the above two inequalities to get the desired with equality for a = b = c

5.For a,b,c R+0 .Prove that -a4 + b4 + c4 abc(

ab +

bc +

ca)

(Source:Nagasaki University 1970)Solution

This problem seems a step tougher to novice. Expanding the RHS will not lead in the correct direction. Letus try to transform this inequality into an equivalent one that for convenient sake looks simpler. One wayof this being done is by taking abc to the LHS -

a4 + b4 + c4 abc(

ab +

bc +

ca)

a3

bc+

b3

ac+

c3

ab

ab +

bc +

ca

a3

bc+

a3

bc+

b3

ca+

c3

ab 4a

b3

ac+

b3

ac+

c3

ba+

a3

cb 4b

c3

ab +

c3

ba +

a3

bc +

b3

ca 4c

the three inequalities are true by AM-GM, add these inequality to get -

a3

bc+

b3

ca+

c3

ab a + b + c

ab +

bc +

ca

notice that the last inequality is true and is equivalent to Problem 2 thus we have proved the inequality withequality for a = b = c

(by M.Ramchandran)aliter: Alternatively another ingenious AM-GM solution will be notice by AM-GM that -

3a4 + 3b4 + 2c4

8 abc

ab

3b4 + 3c4 + 2a4

8 abcbc

3c4 + 3a4 + 2b4

8 abcca

adding these inequalities we get the desired.(by Mathias Tejs)


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Note It is not expected of the reader to get the above two proofs (if he/she is a newbie). Such proofscome due to some strong observation and several wrong tries(like mine.. :P). Both involved splitting theterms into terms with suitable co-efficients which shall come as time goes. So dont get disheartened orawed.6. For a,b,c R+0 such that a + b + c + d = 1.Prove that -

ab + bc + cd 14Solution Write the Inequality using the given condition as -

ab + bc + cd (a + b + c + d)2

4

By AM-GM,

(a + b + c + d)2

4= (

(a + c) + (b + d)

2)2 (a + c)(b + d) = ab + bc + cd + da ab + bc + cd

with the equality for a = b = c, d = 0 or d = c = b, a = 0(by M.Ramchandran)

7. For a,b,c R

+

0 .Prove that -a2 + b2

a + b+

b2 + c2

b + c+

c2 + a2

c + a a + b + c

Solution Note that by AM-GM,

2(x2 + y2) (x + y)2for all non-negative reals x, y thus,

a2 + b2

a + b+

b2 + c2

b + c+

c2 + a2

c + a a + b

2+

b + c

2+

c + a

2= a + b + c

with equality for a = b = c

8. For a,b,c

R

+0 such that a + b + c + d = 1.Prove that -

a2

a + b+

b2

b + c+

c2

c + d+

d2

d + a 1

2

Solution We have -

(a b) + (b c) + (c d) + (d a) = 0

a2 b2a + b

+b2 c2

b + c+

c2 d2c + d

+d2 a2

d + a= 0

a2

a + b+

b2

b + c+

c2

c + d+

d2

d + a=

b2

a + b+

c2

b + c+

d2

c + d+

a2

d + a

thus multiplying the original inequality to be proven by 2 we get that -

a2

+ b2

a + b+ b

2

+ c2

b + c+ c

2

+ d2

c + d+ d

2

+ a2

d + a 1 = a + b + c + d

which can be proved by proceeding similar to the previous question.(by R.Keerthan)

Thus we close this section in the notion that the reader has atleast become familiar with the concepts

that have been explained.


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5. AM-GM TAUTOGRID TECHNIQUE 7

4. Geometric Interpretations

5. AM-GM Tautogrid Technique

While dealing with inequalities sometimes we might be having the denominators as the trouble termsand applying AM-GM in a tricky way we suprisingly are able to solve some rather difficult-looking inequal-ities. I shall demonstrate the method through the following inequalities. It certainly isnt rocket-science just

some common sense which probably the reader might have arrived at with some 5 minutes of genuine thinking.

1. For a,b,c R+.Prove that -a2

b+

b2

c+

c2

a a + b + c

Solution Note that by AM-GM,a2

b+ b 2a

b2

c+ c 2b

c2

a+ a 2c

adding the above inequalities we get the desired with equality for a = b = c

2.For a,b,c R+0 .Prove that -a4 + b4 + c4 abc(

ab +

bc +

ca)

(Source:Nagasaki University1970)

Solution Proceed as in my solution until you get -

a3

bc+

b3

ca+

c3

ab a + b + c

ab +

bc +

ca

The last but one inequality can also be proved in a simple way as follows -

a3

bc+ b + c

3a

b3

ca+ c + a 3b

c3

ab+ a + b 3c

Note: What is the trick? - the idea is the remove the trouble terms and here they are present in thedenominator so we do it by adding suuitably.3. For a,b,c R+0 such that a + b + c + d = 1.Prove that -

a2

a + b+

b2

b + c+

c2

c + d+

d2

d + a 1

2

Solution BY AM-GM,a2

a + b+

a + b

4 a

b2

b + c+

b + c

4 b

c2

c + d+

c + d

4 c

d2

d + a+

d + a

4 d


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adding these we get the desired. with equality for a = b = c = d = 14

(by M.Ramchandran)

4. For a,b,c R+0 such that ab + bc + cd + da = 1.Prove that-

cyclic

a3

b + c + d

1

3Solution By AM-GM,

a3

b + c + d+

b + c + d

18+

1

12 a

2b3

a + c + d+

a + c + d

18+

1

12 b

2c3

b + a + d+

b + a + d

18+

1

12 c

2

d3

b + c + a+

b + c + a

18+

1

12 d

2Add these inequalities to get -

LHS

a + b + c + d

1

3and also frolm AM-GM,

(a + b + c + d)2 4(a + c)(b + d) = 4or

a + b + c + d 2and the conclusion follows. With equallity for a = b = c = d = 1

4.

(by mathlinks user : quykhtn-qa1)

5. For a,b,c R+ such that a + b + c = 2.Prove that-a

b(a + b)+

b

c(b + c)+

c

a(a + c)> 2

(Source:Own Inequality)

Solution By AM-GM,a

b(a + b)+ (a + b)a + ab 3a

b

c(c + b)+ (c + b)b + cb 3b

c

a(a + c)+ (a + c)c + ac 3c

or , cyclic

a

b(a + b) 3(a + b + c) (a + b + c)2 = 6 4 = 2

but the equality case cant occur so the inequality sign becomes strict.

6. Forx,y,z

R

+

such thatxyz

= 1 .Prove that -x3

(1 + y)(1 + z)+

y3

(1 + x)(1 + z)+

z3

(1 + x)(1 + y) 3

4

(Source: IMO Shortlist 1998)Solution By AM-GM

x3

(1 + y)(1 + z)+

1 + y

8+

1 + z

8 3x

4


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6. NESSBITS INEQUALITY 9

=cyclic

x3

(1 + y)(1 + z) 1

4

cyclic

(2x 1) 34

Equality for x = y = z = 1

7. For a, b R+0 such that a + b = 1 Prove that -a2

1 + b+ b

2

1 + a 13

Solution By AM-GM,a2

b + 1+

b + 1

9 2a

3

b2

a + 1+

a + 1

9 2b

3

add these to get the result with equality for a = b =1

26. Nessbits Inequality

This a very famous,well-known and well discussed inequality. Most problems are probably stronger thanthis (that is the job of the proposers - if they keep the questions down to elementary inequality then whatis the fun?) but nevertheless it is a must to know this beautiful inequality-

Theorem 2. For a,b,c R+0 the following inequality holds -a

b + c+

b

c + a+

c

a + b 3

2

ProofNote that for any positive real x, y, z we have by AM-GM -

x

y+

y

z+

z

x 3

Consider the following expressions -

S = ab + c

+ bc + a

+ ca + b

M =b

b + c+

c

c + a+

a

a + b

N =c

b + c+

a

c + a+

b

a + b

we have ofcourse: M + N = 3. According to the Lemma,

M + S =a + b

b + c+

b + c

c + a+

c + a

a + b 3

N + S =a + c

b + c+

b + a

c + a+

c + b

a + b 3

Therefore,

M + N + 2S 3or

2S 3or,

S 32

Practice


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1. Extend the same idea to prove Nessbits inequality for 4 non-negative reals -

a

b + c+

b

c + d+

c

d + a+

d

a + b 4

2. For a,b,c R+0 , Prove that -a2 + 1

b + c+

b2 + 1

c + a+

c2 + 1

a + b 3

(Source : Indian RMO 2006)3. Prove Nessbits inequality using the Tautogrid Technique

(Left to readers)

7. The Reverse Technique

In most inequalities AM-GM can be applied after if for a certain inequality we are to prove - x y, itultimately comes down to the fact that - a2 0 where a2 = x y so factorizations can help. But for factor-izations, one certainly has to be something short of God to be able to prove all inequalities by factorizationsinto squares but.... algebraic manipulation is certainly a powerful tool. In certain inequalities by a trickymanipulation we oberve that solutions are obtained are more easily. In general it should be understood that,for a positive real number which is rational , the value increases as we decrease the denominator and the

value decreases as we increase the denominator and the converse for a negative rational.As always i shall explain this with an example -For a,b,c R+0 , Prove that -

a3

a2 + b2+

b3

b2 + c2+

c3

a2 + c2 a + b + c

2

Solution Seeing the terms a2 + b2, ... , instictively a student does the following -

a3

a2 + b2+

b3

b2 + c2+

c3

a2 + c2 a

3

2ab+

b3

2bc+

c3

2ca=

1

2

a2

b+

b2

c+

c2

a

1

2(a + b + c)

Ok, now is that correct? NO .. ok i guess this was anticipated by you.. but then what was the flaw?read the last lines of the paragraph and the fact that you conceived as elementary has been flawed on by thestudent.

Now an inquisitive student wouldnt move on to a certain different try , He would try to correct his flawto make his idea right. The begining of that important last line said -Positive so that is the cause. Thismeans for applying AM-GM for those terms we need to have the fraction negative, so why not express it assome X Y

a2+b2? Yes that is the central idea behind this useful technique.Now the solution will be -

cyclic

a3

a2 + b2=cyclic

a ab

2

a2 + b2

cyclic

a ab

2

2ab

=cyclic

a b

2

=

a + b + c

2

and Voila! we have a solution and it is correct and it uses AM-GM inequality.

Here the main idea is -at

kat1 + lbt1=

a

k

lk

abt1kat1 + lbt1

This is only a random form. To justify my statement that we can do several types of problems and to get

you used to this technique,The strength and importance of this technique cant be more revealed than the following problems.

Problems1. Let a,b,c R+ , then prove that we have

a3

a2 + ab + b2+

b3

b2 + bc + c2+

c3

c2 + ca + a2 a + b + c

3


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7. THE REVERSE TECHNIQUE 11

Solution

LHS =a3

a2 + ab + b2+

b3

b2 + bc + c2+

c3

c2 + ca + a2

=

cyclic(a ab(a + b)

a2 + ab + b2)

cyclic(a ab(a + b)

3ab) =

a + b + c

3

with equaliy for a = b = c(by M.Ramchandran)

The application of this technique doesnt come by just seeing the solutions given by the Author.. forreally *learning* it, the reader is advised to try the following problems before succumbing to seeing thesolutions.2. Let a,b,c R+ such that , a + b + c = 3 . Prove that

a

1 + b2+

b

1 + c2+

c

1 + a2 3

2

SolutionWe have;

a

1 + b2 = a ab2

1 + b2

. Summing up cyclically we get thatcyc

a

1 + b2= a + b + c

cyc

ab2

1 + b2 a + b + c

cyc

ab2

2b

= 3 12

(ab + bc + ca) 3 32

=3

2since from trivial inequality we have that

(ab + bc + ca) 13

(a + b + c)2 = 3

. Therefore we are done.

Can we extend this problem to four variables? The answer is yes

3. For a,b,c,d R+ such that a + b + c + d = 4 .Show that we havea

1 + b2+

b

1 + c2+

c

1 + d2+

d

1 + a2 2

Solution In the same manner as the previous problem, we have that

a

1 + b2 a ab

2

Summing up we have-

cyc

a1 + b2 a + b + c + d 12 (ab + bc + cd + ad) = 4 12 (a + c)(b + d)

4 12

a + b + c + d

2

2= 4 1

2 4 = 2

Equality holds for a = b = c = d = 1


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4. Let a,b,c R+ satisfy a + b + c = 3. Prove thata2

a + 2b2+

b2

b + 2c2+

c2

c + 2a2

(by Pham Kim Hung)Solution Obviously

a2

a + 2b2= a 2ab2

a + 2b2 a 2ab2

33

a b4 = a 23

a 23 b 23

So we have cyc

a2

a + 2b2 3 2

3

a23 b

23 + b

23 c

23 + c

23 a

23

Hence it suffices to show that

a23 b

23 + b

23 c

23 + c

23 a

23 3

But,

a23 b

23 + b

23 c

23 + c

23 a

23 1

3[ab + ab + 1 + bc + bc + 1 + ca + ca + 1]

= 1 +2

3(ab + bc + ca)

1 +

2

3 3 = 1 + 2 = 3

because(a + b + c)2 3(ab + bc + ca) ab + bc + ca 3

. Hence we finish our proof here.Note that this is also true for ab + bc + ca = 3. However, here I pose a challenge for the readers.

5. For a,b,c R+ such that a + b + c = 3,Prove that -

a2

a + 2b3+

b2

b + 2c3+

c2

c + 2a3 1

Solution Obviouslya2

a + 2b3

= a

2ab3

a + 2b3

a

2

ab3

33

a b2

= a 23

b3

a2

Hence we have cyc

a2

a + 2b3 a + b + c 2

3

b3

a2 + c3

b2 + a3

c2

a + b + c 29

[b(a + a + 1) + c(b + b + 1) + a(c + c + 1)]

= 3 29{2(ab + bc + ca) + 3} 3 2

9{6 + 3} = 1

The last inequality is true by AM-GM, and since we have

ab + bc + ca 13

(a + b + c)2 = 3.

Equality occurs if and only if a = b = c = 1.

6. For a,b,c R+ such that a + b + c = 3, Prove that -

a + 1

b2 + 1+

b + 1

c2 + 1+

c + a

a2 + 1 3


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7. THE REVERSE TECHNIQUE 13

Solution Sincea + 1

b2 + 1= a + 1 b

2(a + 1)

b2 + 1 a + 1 b

2(a + 1)

2b= a + 1 ab

2 b

2. Therefore we have

cyc

a + 1

b2 + 1 3 + a + b + c

2 1

2(ab + bc + ca)

3 + 32

32

= 3

Equality for a = b = c = 1Hence proved.

7. For a,b,c R+ such that a2 + b2 + c2 = 3, prove that

1

a3 + 2+

1

b3 + 2+

1

c3 + 2 1

(Pham Kim Hung)Solution Note that

cyc

1a3 + 2

=cyc

12

1 a

3

a3 + 2

32

12

cyc

a3

3a=

3

2 1

2= 1

8. For a, b,c > 0. Prove that -

a3

a + b+

b3

b + c+

c3

c + a 1

2

a2 + b2 + c2

Solution

a3

a + b= a

2 a2b

a + b= a2 + b2 + c2

a2ba + b

a2 + b2 + c2 1

2 a

2bab

= a2 + b2 + c2 1

2(a

ab) a2 + b2 + c2

a4

(a + b)

= a2 + b2 + c2 14

(a2 + b2 + c2 + ab + bc + ca) a2 + b2 + c2 12

(a2 + b2 + c2)

=1

2(a2 + b2 + c2)

Therefore we are done. Equality occurs if and only if a = b = c.

9. Let a,b,c R+ that sum up to 3. Prove that we always have -

11 + 2b2c

+ 11 + 2c2a

+ 11 + 2a2b

1Solution Note that

1

1 + 2b2c= 1 2b

2c

1 + 2b2c 1 2

3

b2c3

b4c2

= 1 23

3

b2c 1 29

(2b + c)


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Therefore we have that cyc

1

1 + 2b2c 3 2

9

cyc

(2b + c)

= 3

2

9 3(a + b + c) = 3

2 = 1

Therefore we are done. Equality occurs if and only if a = b = c = 1.

Practice Problems

10. For a,b,c,d R+ such that a + b + c + d = 1.Prove that -a2

a + b+

b2

b + c+

c2

c + d+

d2

d + a 1

2

11. Given a,b,c,d R+, show that we havea4

a3 + 2b3+

b4

b3 + 2c3+

c4

c3 + 2d3+

d4

d3 + 2a3 a + b + c + d

3

12. For a,b,c R+ such that a + b + c = 3; show that we haveab

b3 + 1+

bc

c3 + 1+

ca

a3 + 1 3

2

(by Gibbenergy)13. For given four positives a,b,c,d with sum 4; show that

a

1 + b2c+

b

1 + c2a+

c

1 + d2a+

d

1 + a2b 1

(by Pham Kim Hung)15.Let a,b,c,d > 0 satisfy a + b + c + d = 4; show that

1 + ab

1 + b2c2+

1 + bc

1 + c2d2+

1 + cd

1 + d2a2+

1 + ad

1 + a2b2 4

(by Pham Kim Hung)16. For all a,b,c,d R+ satisfying a + b + c + d = 4, Prove that we have

a + 1

b2 + 1+

b + 1

c2 + 1+

c + 1

d2 + 1+

d + 1

a2 + 1 4

17. For a,b,c,d R+ with sum 4. Prove that -1

a2 + 1+

1

b2 + 1+

1

c2 + 1+

1

d2 + 1 2


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8. THE WEIGHTED AM-GM INEQUALITY 15

8. The Weighted AM-GM Inequality

Theorem 3. Suppose that a1, a2, a3, ...., an are positive real numbers.If n nonegative real numbersx1, x2, x3..,xn have sum 1 then,

x1a1 + x2a2 + .... + xnan ax11 ax22 ...axnn

In a way similar to the second proof provided for AM-GM. We have to prove that ifx, y

0, x

+y

= 1and a,b > 0ax + by axby

Consider rational numbers x, y then take a limit. Certainly ifx,y, are rational numbers then - x = mm+n

, y =n

m+n, m, n N, the problem is true according to AM-GM inequality -

ma + nb (m + n)a mm+n b nm+n = ax + by axbyIf x, y are real numbers,there exist two sequences of rational numbers (rn)n0 and (sn)n0 for which rn x, sn y, rn + sn = 1.Certainly

arn + bsn arnbsnor

arn + b(1 rn) arnb1rnTaking the limit when n

, we have ax + by

axby

Problems

1. Let a,b,c be the sidelengts of a triangle.Prove that -

(a + b c)a(b + c a)b(c + a b)c aabbccSolution Applying the weighted AM-GM inequality, we conclude that

a + b ca

a

b + c ab

b

c + b ac

c 1a+b+c

1a + b + c

a a + b c

a+ b b + c a

b+ c c + a b

c

= 1

Or equivalenty,

(a + b c)a

(b + c a)b

(c + a b)c

aa

bb

cc

Equality occurs for a = b = c

2. Let a,b,c R+ such that abc = 1.Prove that -ab+c bc+a ca+b 1

(Source : INMO 2001)Solution 1 Write the LHS as - (ab)c(bc)a(ca)b.So we have to prove that -

(ab)c(bc)a(ca)b 1By Weighted AM-GM inequality we have -

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

1

a+b+c 1a + b + c

(a.bc + b.ca + c.ab) =3

a + b + c 1

(By Ramchandran)Solution 2 WLOG:c b a abc = 1, so c 1,so ab 1 and a 1 Now:

ab+c bc+a ca+b= (abc)a + bacabcb

= acabcb

= (ab)cbaba


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1cb1ba

= 1

Therefore ab+c bc+a ca+b 1(by mathlinks user: rem)

Solution 3 Note that -ab+c bc+a ca+b = 1

aabbcc 1

the last inequality is true because for a,b,c individually greater or lesser than 1 we have -

aa a, bb b, cc c(by mathlinks user:Maharjun)

3. Let a,b,c R+0 , Prove that -

4(a + b + c) 3(a + (ab) 12 + (abc) 13 )Solution

3a + (a

4+ b) + (

a

2+ 2b) + (

a

4+ b + 4c) 3a +

ab + 2

ab + 3

3

abc

(by Aravind Srinivas)

9. Method Of Balancing Co-efficients by AM-GM

In most inequalities we have to group terms suitably so that classical inequalities like AM-GM , C-S etc..can be aplied to get the result.This is not as easy as it looks, it requires proper terms and proper grouping.We usually need some additional variables to solve the equations for finding out the original variables.Thisis the Method of Balancing co-efficients. In this chapter we shall see the method using AM-GM Inequality.

Let me demonstrate it with a simple example -

1. Ifx, y, z R+

such that xy + yz + zx = 1 ,then find the minimum of the following expression-

k(x2 + y2) + z2

Solution Lets experiment with some values of k shall we? Let k = 10, so we are now required to find theminimum of this non-symmetric expression-

10(x2 + y2) + z2

How do we apply a classical basic inequality like AM-GM for this? Well it does seem horrendously difficult,so lets take a sneak-peek at the magical solution? By AM-GM we have the following inequalities,

2x2 + 2y2 4xy

8x2 + 12 z2 4yz

8y2 +1

2z2 4zx

and summing up these we have,

10(x2 + y2) + z2 4(xy + yz + zx) = 4


23/37

9. METHOD OF BALANCING CO-EFFICIENTS BY AM-GM 17

Equality holds for

x = y

4x = z

4y = z

= x = y =1

3

z =4

3

Hurray ! We did it!!(dont get envious i like you was flabbergasted at this cause it aint mine) how cana guy arrive at this?? why not pair up 1 and 9 , 5 and 5 or something? All answers shall be revealed in thefollowing lines. back to our general problem, Lets choose some l k , then apply AM-GM this way,

lx2 + ly2 2lxy

(k l)y2 + 12

z2 yx

2(k l)

(k l)x2 + 12

z2 xz2(k l)summing up we get this -

k(x2 + y2) + z2 2lxy + (yz + zx)

2(k)now we have a condition given - xy + yz + zx = 1 so for obtaining a numerical value we have to have theco-efficients in the RHS the same and without a variable hopefully. so intuitively lets just equate them-2l =

2(k l) and solving this we obtain that

l =1 + 1 + 8k

4

and ofcourse substituting this in the equation we get the minimal value we are looking for to be -

1 + 1 + 8k2

so we observe that there is a unique pair of integers - l, k

l that show us the way by AM-GM. Now that is

the reason we were seeing the use of 8,2 and not 3,7 etc.. sure enough you can check the credibility of thepairings now!

Note

2. Let x, y, z , t be real numbers satisfying xy + yz + zt + tx = 1. Find the minimum of the expression -

5x2 + 4y2 + 5z2 + t2

Solution Here k = 5, so we chooose l 5,lx2 + 2y2 2

2lxy

2y2 + lz2 2

2lyz

(5 l)z2 + 12

t2 2(5 l)txSumming up these results, We conclude that

5x2 + 4y2 + 5z2 + t2 2

2l(xy + tz) +

2(5 l)(zt + tx)The condition xy + yz + zt + tx = 1 suggests us to choose a number l(0 l 5) such that -

2

2l =

2(5 l). A simple calculation yields l = 1, thus the minimum of 5x2 + 4y2 + 5z2 is 2

2


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Note Using the same method, solve the following problem :Let x, y, z , t be arbirary real numbers. Prove that -

x2 + ky2 + z2 + lt2

2kl

k + l

12

(xy + yz + zt + tx)

3. Let x, y, z be positive real numbers with sum 3. Find the minimum of the expressionx2 + y2 + z3

(Pham Kim Hung)Solution Let a and b be teo positive real numbers.. Then,by AM-GM inequality we have -

x2 + a2 2axy2 + a2 2ay

z3 + b2 + b2 3zb2Combining these we have x2 + y2 + z3 + 2(a2 + b3) 2a(x + y) + 3b2z with equality for x = y = a andz = b.In this case, we could have 2a + b = x + y + z = 3(). Moreover, in order for 2a(x + y) + 3b2z to berepresented as x + y + z, we must have 2a = 3b2().Accordin to () and () we can find out that,

b = 1 + 376

, a = 3 b2

= 19 3712

Therfore the minimum ofx2 + y2 + z3 is 6a 2(a2 + b3) where a, b are as determined.The proof is completed.

4. For x,y,z R+ such that xy + yz + zx = 1 , Prove that - 15x2 + 7y2 + 3z2 6(Own Inequality)

Solution Rewrite the Inequality as -

5x2 +7

3y2 + z2 2

By AM-GM we have ,

x2 + y2 2xy

4x2

+

1

4 z2

2xz4

3y2 +

3

4z2 2xz

now adding these we get the desired.

10. Quasiliearisation

This is a very intrigueing idea due to Russian problem proposer - Fedor PetrovWe know by AM-GM that ,

2ab a2 + b2Introduce a parameter and get the following:

2ab

ta2 +

b2

tThen read the last inequality from the other point: for any positive a, b there exist positive t such that

2ab = ta2 +b2

t(t =

b

a)

Or, we may write:

2ab = min(ta2 +b2

t)


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11. EQUIVALENT SUMMATION TECHNIQUE 19

How may this observation help? Put

a =

(x2i ), b =

(y2i )

Then for appropriate t we have:

2

a

b = ta +b

t

= (tx2i +y2i

t

)

(2xiyi)

So, we get the Cauchy-Schwarz Inequality. A lot of other inequalities also may be proved by this ideaProblem Prove that for any four nonnegative reals a,b,c,d the following inequality holds-

(ab)13 + (cd)

13 ((a + c + b)(a + c + d)) 13

(Source:Proposed at 239 Lyceum Traditional Olympiad)(Author : Fedor Petrov)

We have

3(AB)13 Ax + By + 1

xy

And for any positive A and B there exist appropriate x and y ,for which equality holds

x =(AB)

13

A, y =

(AB)13

B

LetA = (a + c + b), B = (a + c + d)

in terms of the problem. For some positive x, y we have -

3(AB)13 = Ax + By +

1

xy= (a + c + b)x + (a + c + d)y +

1

xy=

(a+c+b)x+(a+c+d)y+1

x(x + y)+

1

y(x + y)=

a(x+y)+bx+1

x(x + y)

+

c(x+y)+dy+1

y(x + y)

3(ab) 13 +3(cd) 13By AM-GM and we are done!

11. Equivalent Summation Technique

This is an interesting technique that helps us to solve elegantly many problems. It involves finding asuitable equivalent summation and using it prove the inequality given. Let me demonstrate it through thefollowing already discussed example -1. Prove that for a,b,c R+ -

a3

a2 + ab + b2+

b3

b2 + bc + c2+

c3

c2 + ca + a2 a + b + c

3

Solution 1

a3 b3a2 + ab + b2

= a b =cyc

a3

a2 + ab + b2=cyc

b3

a2 + ab + b2=

1

2

cyc

a3 + b3

a2 + ab + b2 a + b + c

3

But there exists another nice solution using the reverse technique. The solution runs as follows:

cyc

a3

a2 + ab + b2=

cyca

ab(a + b)

a2 + ab + b2 a + b + c cyc

ab(a + b)

3ab

Since we have a2 + ab + b2 3ab from AM-GM. Therefore we have thatcyc

a3

a2 + ab + b2 a + b + c 2

3(a + b + c) =

a + b + c

3

Hence proved. Equality occurs if and only if a = b = c.


26/37


Solution 2 By the same idea,

LHS =1

2

a3 + b3a2 + b2 + ab

. But

ab a2 + b2

2

and

2(a3 + b3) (a + b)(a2 + b2)Therefore

LHS a + b + c3

2. For a,b,c R+ with sum 1,prove that -

abab + bc

+bc

bc + ca+

caca + ab

12

(MOSP 2007 3.2)Solution

abab + bc

+bc

bc + ca+

caca + ab

12

abab + bc

2ab

ab +

bc

just need to prove

ab

ab +

bc a + b + c

2

note that abab +

bc

= bc

ab +

bc

this is true because :

ab bcab +

bc

= (ab bc)(ab bc)

ab bc =

(

ab

bc) = 0

so equivalent to ab + bcab +

bc

a + b + c

and this is equivalent to -

ab(

c +

a)(

c +

b)(

a

b)2 0

and hence proved.(by mathlinks user : kuing)


27/37

12. THE G FUNCTION 21

12. The G function

This is a beautiful idea for which credit goes to inequality solver - Pham Kim Hung (hungkhtn).

Definition 3. For a,b,c R+ , we define,G(a,b,c) =

a

b

+b

c

+c

a 3

It is trivial to oberve by AM-GM that always -

G(a,b,c) 0Some nice properties have been found and some tough inequalities have been solved by this idea.Pham Kim Hungs Nice Factorisation

G(a,b,c) =a

b+

b

c+

c

a 3 =

a

b+

b

a 2

+

b

c+

c

a b

a 1

=(a b)2

ab+

(b c)(a c)ca

this factorization plays an important role in many proofsNote

The inequality - G(a,b,c)

0 is a cyclic inequality and thus no pair-wise order can be assumed.

I shall present some properties of this important function.Properties

1. For a,b,c,k R+ we have -G(a,b,c) G(a + k, b + k, c + k)

Proof WLOG : c = min{a,b,c} We have ,

G(a,b,c) =(a b)2

ab+

(b c)(a c)ca

So it is enough to prove that -

(a b)2ab

+(b c)(a c)

ca (a b)

2

(a + k)(b + k)+

(b c)(a c)(c + k)(a + k)

This is true as k > 0 and by assumption - (a

c)(b

c)

0. Hence proved with equality for a = b = c

2. For a,b,c,k R+ we have -

G(a,b,c) G(a + b, b + c, c + a)Proof We only have to show that -

(a b)2ab

+(b c)(a c)

ca (a b)

2

(a + c)(b + c)+

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

It is trivial to see that the inequality is true.Note The following problem (equivalent to the property discusses above) was asked in the Mathlinks

contest 2003 -a

b+

b

c+

c

a a + b

b + c+

b + c

c + a+

c + a

a + b

The following properties(same conditions as above)can be solved using the same method -3. For a,b,c,k R+ we have -

G

(a b)2, (b c)2, (c a)2 2(by Darij Grinberg)

4. For a,b,c,k R+ we have -G(a2, b2, c2) 4G(a,b,c)


28/37


(by M.Ramchandran)5. Let a,b,c,k R+, then If and only if the sum of any two of {a,b,c} is less than 2

G(ab, bc, ac)

G(a,b,c)

(by M.Ramchandran)6. For a,b,c,k R+ and let a b c then,

G(a2

bc,

b2

ca,

c2

ab) G(a,b,c)

(by M.Ramchandran)7. For a,b,c,k R+ and let k max{a2, b2, c2} then,

G(a,b,c) G(a2 + k, b2 + k, c2 + k)(Pham Kim Hung)

8. For a,b,c,k R+ we have -G(a3, b3, c3) 3G(a2, b2, c2)

(by M.Ramchandran)9. For a,b,c,k R+ we have -

G(a2, b2, c2) G(a,c,b)

(by M.Ramchandran)If more properties are invited to be shared with the author by e-mail.

13. Problem Set

This section consists of problems a step more difficult then the problems already discussed.1. Let a,b,c R+ such that, a + b + c = 3.Prove that -

a +

b +

c ab + bc + ca

(Source : Russian MO 2004)Solution By AM-GM,

a2 +

a +

a 3ab2 +

b +

b 3b

c2 +

c +

c 3cThus,by adding the above and using a + b + c = 3 ,

a2 + b2 + c2 + 2(

a +

b +

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

=

2(

a +

b +

c)

2(ab + bc + ca)

= a + b + c ab + bc + caWith equality for a = b = c = 1

2.Let x, y, z R+.Prove that -

1 +x

y

1 +

y

z

1 +

z

x

2 + 2(x + y + z)

(xyz)13


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13. PROBLEM SET 23

(Source : APMO 1998)Solution After expanding, the inequality reduces to -

x

y+

y

z+

z

x x + y + z

(xyz)13

For the proof of this inequality see Lemma . Equality for x = y = z

3.Let a,b,c be positive real numbers.Prove that1

a3 + abc + b3+

1

b3 + abc + c3+

1

c3 + abc + a3 1

abc

(USA MO 1998)Solution Using the lemma : a3 + b3 ab(a + b)

abc

a3 + b3 + abc abc

ab(a + b) + abc=

c

a + b + c

Construction two more similar inequalities and adding we get the desired result with equality for a = b = c

4. Ifx1, x2, ...xn R+ such that -1

1 + x1+

1

1 + x2+ ... +

1

1 + xn= 1

, Prove that -x1x2x3....xn (n 1)n

Solution The condition is equivalent to -

1

1 + x1+

1

1 + x2+ ... +

1

1 + xn1=

xn

1 + xn

Using AM-GM inequality for all the terms in the LHS we get -

xn

1 + xn n 1

((1 + x1)(1 + x2)(1 + x3)...(1 + xn))1

n1

Similarly constructing n more inequalities and multiplying all we get the desired. Equality for xi = n 1for i {1, 2, 3, ...n}

5. Suppose that x, y, z R+ and x5 + y5 + z5 = 3.Prove that -x4

y3+

y4

z3+

z4

x3 3

Solution Notice that -

(x5 + y5 + z5)2 = x10 + 2x5y5 + y10 + 2y5z5 + z10 + 2z5x5 = 9

This suggests the use of AM-GM in this way -

10 x4

y3+ 6x5y5 + 3x10 19x 10019

Adding all the cyclic results we get,

10(x4

y3

+y4

z3

+z4

x3

) + 3(x5 + y5 + z5)2

19x10019 + y 10019 + z 10019

Therfore it is enough if we prove -

(x10019 + y

10019 + z

10019 x5 + y5 + z5

which is true by AM-GM because -

3 + 19cyclic

=cyclic

(1 + 19x10019 ) 20

cyclic

x5


30/37



6. Suppose that x, y, z R+ and x4 + y4 + z4 = 3.Prove that -x256 + y

256 + z

256 3

(Own Inequality)

Solution The inequality is equivalent to -

3 + 24cyclic

x256 25x4

which is true by AM-GM : cyclic

(1 + 24cyclic

x256 ) 25

cyclic

x4


7. Let a,b,c R+ such that abc = 1.Prove thata + b

a + 1+

b + c

b + 1+

c + a

c + 1 3

(Source : Mathlinks Contest)Solution After applying AM-GM to the left hand side we get -

(a + b)(b + c)(c + a) (a + 1)(b + 1)(c + 1)and since abc = 1 it is equivalent to -

ab(a + b) + bc(b + c) + ca(c + a) a + b + c + ab + bc + caBy AM-GM,

2LHS +cyclic

ab cyclic

a2b + a2b + a2c + a2c + bc

5 cyclic

[a5 (abc) 13 ] 15 = 5cyclic

a

2LHS + cyclic a = cyclic a2b + a2b + b2a + b2a + c 4 cyclic[(ab)

5 .abc] 15

Therefore,

4LHS +cyclic

ab +cyclic

a 5cyclic

a + 5cyclic

ab

= 4LHS 4

cyclic

a +cyclic

ab

Hence Proved. Equality for a = b = c = 1

8. Let a,b,c,d R+0 such that a + b + c + d = 4. Prove that -a2 + b2 + c2 + d2 4 4(a 1)(b 1)(c 1)(d 1)

(Pham Kim Hung)

Solution By AM-GM ,

a2 + b2 + c2 + d2 4 = (a 1)2 + (b 1)2 + (c 1)2 + (d 1)2 4

|(a 1)(b 1)(c 1)(d 1)|If the RHS of the question is negative, then the question is meaningless . So we only have to consider thecase when - a b 1 c d Since a + b 4 and c, d 1 and by AM-GM,

(1 c)(1 d) 1


31/37

13. PROBLEM SET 25

(a 1)(b 1) 14

(a + b 2)2 1Therefore -

(a 1)(b 1)(1 c)(1 d) 1and the conclusion follows. Equality for a = b = c = d = 1, a = b = 2, c = d = 0 and cylic permutations.

9. Let a,b,c R+0 and a + b + c = 3 . Prove that -a

1 + b3 + b

1 + c3 + c

1 + a3 5(Pham Kim Hung)

Solution We know by AM-GM that,cyclic

a

1 + b3 =cyclic

a

(1 + b)(1 b + b2) cyclic

1

2 a(2 + b2)

Thus we are left to prove that -

ab2 + bc2 + ca2 4WLOG : Let b be the middle number in {a,b,c} So we have ,

a(b

a)(b

c)

0

= ab2 + a2c abc + a2bThus is it is enough if we prove that -

abc + a2b + bc2 4 b(a2 + ac + c2) 4By AM-GM inequality,

b(a2 + ac + c2) b (a + c)2 = 4 b a + c2

a + c2

4

a + b + c

3

3= 4

This finished our proof with equality for a = 1, b = 2, c = 0 and cyclic permutations.

10. Let a,b,c R+.Prove that -a3

b2+

b3

c2+

c3

a2 a2

b+

b2

c+

c2

a(Source : JBMO Shortlist 2002)

Solution 1 Note that,a3

b2 a

2

b+ a b

or,

a3 a2b + ab2 b3which is true by Lemma . Add all cyclic results to get the desired

(by mathlinks user : limes123)Solution 2 By AM-GM, we know that

cyc

a3

b2+ a

cyc

2a2

b

We shall prove , cyc

2a2

b a b c

cyc

a2

b

or, cyc

a2

b a + b + c


32/37


It is true by AM-GM, a2b

+ b

cyc

2a

Hence Proved.(by Johan Gunardi)

Solution 3Lemma For a, b R+,

a3 + b3 ab(a + b)We have,

a3

b2+

b3

c2+

c3

a2

+ (a + b + c) =

a3b2

+ b

= a3 + b3

b2 ab (a + b)

b2=a2

b+ a

= a3

b2+

b3

c2+

c3

a2 a

2

b+

b2

c+

c2

a(by mathlinks user : trungk42sp)

Solution 4 Note by AM-GM that,

14a3

b2+ 3

b3

c2+ 2

c3

a2 19a38

b19=

a2

b(by mathlinks administrator : nsato)

11. For a,b,c R+.Prove that -

a6

b3+

b6

c3+

c6

a3 b

4

a+

c4

b+

a4

c(Vascile Cirtoaje)

Solution Note that by AM-GM we have -

3 a6

b3=

(b6

c3+

b6

c3+

c6

a3) 3

b4a

(by mathlinks user : karis)

12. For a,b,c,d R+.Prove that -a14

b7+

b14

c7+

c14

d7+

d14

a7 b

8

a+

c8

b+

d8

c+

a8

d

(Vascile Cirtoaje)Solution Note that by AM-GM we have -

7 a14

b7=

(4b14

c7+ 2

c14

d7+

d14

a7) 7

b8a

(by mathlinks user : karis)

13. Let a,b,c R+ such that they are all pairwise distinct . Prove that -

a + ba b

+b + c

b c+

c + a

c a > 1(Source : Iranian National Olympiad (3rd Round) 2007)

Solution Set ,a + b

a b = xb + c

b c = y


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13. PROBLEM SET 27a + c

c a = z

= xy + yz + zx = 1By AM-GM we have,

(x + y + z)2

3(xy + yz + zx) = 3 =

|x + y + z

|

3 >

(by mathlinks user : Phm Thnh Quang)

14.Prove the following inequality for all a,b,c R+0 -a

b+

b

c+

c

a+

3(abc)13

a + b + c 4

Solution Lemma For a,b,c R+.a

b+

b

c+

c

a a + b + c

(abc)13

Proof By AM-GM we have -

cyclic

2 ab

+b

c 3

cyclic

a2

bc13

= 3 cyclic

a

(abc)13

we have to prove that -a + b + c

(abc)13

+3(abc)

13

a + b + c 4

By AM-GM,

a + b + c

3(abc)13

+a + b + c

3(abc)13

+a + b + c

3(abc)13

+3(abc)

13

a + b + c 4

a + b + c

3(abc)13

4

Hence proved with equality for a = b = c(by mathlinks user : enndb0x)

15. Let a,b,c R+ such that abc = 1.Prove that -

a2 + b2 + c2 + 9(ab + bc + ca)

10(a + b + c)

Solution This beautiful problem has many solutions - mostly involving some high-level methods likeuvw method, mixing variables etc..But, the following extra-ordinary generalisation was given by an Indian- Aakansh Gupta -

cyc

a2 + kcyc

ab (k + 1)cyc

a k R

ProofWe have

cyc

a 3 andcyc

ab 3

(cyc a)2

+ (cyc ab)2 3cyc a + 3cyc ab

cyc

a2 +cyc

a2b2 cyc

a +cyc

ab

(cyc

a2 + kcyc

ab) + (cyc

a2b2 + kcyc

a) (k + 1)cyc

a + (k + 1)cyc

ab

cyc

a2 + kcyc

ab (k + 1)cyc

a


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Or, cyc

a2b2 + kcyc

a (k + 1)cyc

ab

If first one occurs we are done and if the second one occurs then replace a by 1a

; b by 1b

; c by 1c

and weget the same expression as the first one and thus we have proved !!

16.Prove that for a,b,c R+ -2a

a + b+

2b

b + c+

2c

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

(a2 + b2 + c2) (a2b2 + b2c2 + c2a2)

Solution 2b

b + c 4b

3b + c2c

c + a 4c

3c + a2a

a + b 4a

3a + b

So that, 2a

a + b+

2b

b + c+

2c

c + a 4

a

3a + b+

b

3b + c+

c

3c + a

From Titus Lemma and the following well-known inequality a2 + b2 + c2 ab + bc + ca

a

3a + b+

b

3b + c+

c

3c + a=

a2

3a2 + ab+

b2

3b2 + bc+

c2

3c2 + ca

(a + b + c)2

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

2

4 (a2 + b2 + c2)

Therefore,

2a

a + b+

2bb + c

+

2cc + a

(a + b + c)2

a2 + b2 + c2

We also have, (just prove analogously)2a

a + b+

2b

b + c+

2c

c + a=

2ac

ac + bc+

2ba

ba + ca+

2cb

cb + ab

(ab + bc + ca)2

a2b2 + b2c2 + c2a2

So that,

2a

a + b+

2b

b + c+

2c

c + a 1

2

(a + b + c)

2

a2 + b2 + c2+

(ab + bc + ca)2

a2b2 + b2c2 + c2a2

(a + b + c) (ab + bc + ca)(a2 + b2 + c2) (a2b2 + b2c2 + c2a2)

(from AM-GM inequality)The proof is completed. Equality holds if and only if a = b = c

(by mathlinks user : leviethai)


35/37


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by applying AM-GM to the denominators in the RHS we get -

RHS cyclic

ak32 b

12

Thus, it is enough if we prove that -

2(ak1 + bk1 + ck1)

3 + ak32 b

12 + bk

32 c

12 + ck

32 a

12

This follows directly from AM-GM as -

ak1 + bk1 + ck1 3 (abc)k13 = 3And also ,

(2k 3)ak1 + bk1 (2k 2)ak 32 b 12(2k 3)bk1 + ck1 (2k 2)bk 32 c 12(2k 3)ck1 + ak1 (2k 2)ck 32 a 12

Adding the above we get the desired.Equality for a = b = c = 1(By Pham Kim Hung)

19. Prove that for a,b,c R+, we have -

1 + 3ab + bc + ca

6a + b + c

(Source : Macedonia Team Selection Test 2007)Solution The inequality is equivalent to -

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

ab + bc + ca 6

By AM-GM inequality we have,

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

ab + bc + ca 2

3(a + b + c)2

ab + bc + ca 6

Equality for a = b = c = 1(By Vo Danh)

20. Let a,b,c 0 such that abc 1 . Prove that -a +

1

a + 1

b +

1

1 + b

c +

1

1 + c

27

8

(Source : Ukraine Mathematical Fes-tival 2007)

Solution By AM-GM we have - ,a + 1

4+

1

1 + a 1

and ,3a

4+

3

4 3

2 a

Adding the two inequalities we get -

a +1

1 + a 3

2 a

Obtaining similarly all the cyclic inequalities and multiplying them we get -a +

1

a + 1

b +

1

1 + b

c +

1

1 + c

27

8

abc 278

Equality for a = b = c = 1


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13. PROBLEM SET 31

(by Nguyen Dung TN)

My Inequality Project (1)

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Transcript of My Inequality Project (1)