Showing posts with label IMO QUESTION & ANSWER. Show all posts
Showing posts with label IMO QUESTION & ANSWER. Show all posts

Saturday, June 27, 2020

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 aand bi,we have
Condition:
Equality holds if and only if  ai = kbi, 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.....aand b1 ,b2 ,b3....bis
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 and its permutations

This is how we prove.

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