$2009$ USA TST Inequality

For discussing Olympiad Level Algebra (and Inequality) problems
rah4927
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Joined:Sat Feb 07, 2015 9:47 pm
$2009$ USA TST Inequality

Unread post by rah4927 » Sun Feb 19, 2017 11:16 am

Prove that for positive real numbers $x$, $y$, $z$, $$x^3(y^2+z^2)^2 + y^3(z^2+x^2)^2+z^3(x^2+y^2)^2 \geq xyz\left[xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\right]$$

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Atonu Roy Chowdhury
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Joined:Fri Aug 05, 2016 7:57 pm
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Re: $2009$ USA TST Inequality

Unread post by Atonu Roy Chowdhury » Thu Mar 23, 2017 11:36 pm

Just substitute $x = 1/a$, $y = 1/b$ and $z = 1/c$.
The rest is so cool.
This was freedom. Losing all hope was freedom.

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