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Saturday, January 16, 2016

IMO Inequality Problem

Let a,b,c be real numbers greater than 2 such that \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1.

Prove that (a-2)(b-2)(c-2)\le 1.

My solution:

Note that

(a−2)(b−2)(c−2)

=abc\left(\dfrac{a-2}{a}\right)\left(\dfrac{b-2}{b}\right)\left(\dfrac{c-2}{c}\right)

=abc\left(1-\dfrac{2}{a}\right)\left(1-\dfrac{2}{b}\right)\left(1-\dfrac{2}{c}\right)

=abc\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{2}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{2}{b}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{2}{c}\right)

=abc\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)

We now use the famous identity that says for all real and positive x,\,y and z, we have

xyz\ge (x+y-z)(x+z-y)(y+z-x)

In our case, we have x=\dfrac{1}{a},\,y=\dfrac{1}{b},\,z=\dfrac{1}{c} and so we get

\dfrac{1}{abc}\ge \left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}\right)\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a}\right)

i.e.

abc\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}\right)\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a}\right)\le 1

The proof is then follows.

Equality occurs when a=b=c=3.


2 comments:

  1. Aww...thank you Michelle for your nice compliment, I appreciate it!

    ReplyDelete