IMO shortlist 1996.

Posts: 791
Joined: Aug 4, 2011

by hemangsarkar, Sep 1, 2012, 6:05 PM

Q) consider three positive real numbers, $x,y,z$ whose product is $1$ . prove that -

$\sum_{cyclic} \frac{xy}{x^5 + xy + y^5} \le 1$

my solution :

it is worth noting that
$\frac{xy}{x^5 + xy + y^5} \le \frac{z}{x + y + z}$
since on solving this we get,

$x^2y + y^2x \le x^5z + y^5z$

plug in $z = \frac{1}{xy}$

$x^2y + y^2x \le \frac{x^4}{y} + \frac{y^4}{x}$

or, $x^4y + y^4x \le x^5 + y^5$

which is easy to prove by muirhead's.
similarly we can prove that -

$\frac{yz}{y^5 + yz + z^5} \le \frac{z}{x+y+z}$

and $\frac{zx}{z^5 + xz + x^5} \le \frac{y}{x+y+z}$
so we have

$\sum_{cyclic} \frac{xy}{x^5 + xy + y^5} \le \sum_{cyclic} \frac{x}{x+y+z} = 1$

hence proved.

equality holds if $(x,y,z) = (1,1,1)$ .