A QUESTION FROM IMO 1995 | TITU'S LEMMA | CAUCHY-SCHWARZ INEQUALITY

INTRODUCTION:

In this session we will discuss about a question which had appeared in IMO 1995. 

So before staring the session i will discuss about ' Titu's Lemma ' and  ' Cauchy-Schwarz Inequality '.

Cauchy-Schwarz Inequality :

 The Cauchy-Schwarz inequality also known as Cauchy-Bunyakovsky-Schwarz inequality states that for all sequence of real number a_i and b_i,we have

Condition:

Equality holds if and only if  a_i = kb_i, for non-zero constant K € R.

TITU'S LEMMA:

Titu's lemma is a direct consequence of cauchy-Schwarz inequality .

It is also known as Engle's Form, or Sedrakyan's inequality.

      Titu's Lemma states that for positive real number a₁,a₂,a₃.....a_n and b_{1 ,}b_{2 ,}b_3....b_n is

This form is specially helpful when the inequality involvs fractions where the numeretor is a perfect square.

Then it becomes

           

QNs.

Let  a,b,c are positive real numbers such that a.b.c = 1.

Prove that

ANS.

There are two ways to solve the question 

1.By Titu's Lemma

2. By AM-GM inequality

Method-1:

We have already discussed the inequality at the top.

Substitute a= 1/x  , b= 1/y , c=1/z

by the use of Cauchy-Schwarz inequality 

And the Titu's lemma

we know that a.b.c =1 so ,the  result will be 3/2.

Method-2:

Consider (a+b) , (b+c) ,(c+a).Then by the AM - GM inequality, we have the first set of inequalities.

Multiplying altogether 

But  abc=1 ,then

       (a+b)(b+c)(c+a) >=   8

By the power mean inequality ( AP > GP >HP), and considering the arithmetic and harmonic means of a^3(b+c) and its permutations

This is how we prove.

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