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.
Well done Belle. Good job! :-))
ReplyDeleteAww...thank you Michelle for your nice compliment, I appreciate it!
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