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Problem A. 487. (September 2009)

A. 487. Let x, y, z be positive numbers satisfying xyz\ge1. Prove that \frac{x}{x^3+y^2+z}+ \frac{y}{y^3+z^2+x}+ \frac{z}{z^3+x^2+y}\le
1.

Proposed by Tuan Le, Anaheim, California, USA

(5 pont)

Deadline expired on October 12, 2009.


Solution. By the Cauchy-Schwarz inequality,


\left(\big(x^{3/2}\big)+y^2+\big(\sqrt{z}\big)^2\right) 
\left(\big(1/\sqrt{x}\big)^2+1^2+\big(\sqrt{z}\big)^2\right) \ge
\left(\big(1/\sqrt{x}\big)\cdot\big(x^{3/2}\big)+
y\cdot1 + \sqrt{z}\cdot\sqrt{z} \right)^2,


(x^3+y^2+z)\left(\frac1x+1+z\right) \ge (x+y+z)^2.

Therefore


\frac{x}{x^3+y^2+z} \le \frac{1+x+xz}{(x+y+z)^2},

and similarly


\frac{y}{y^3+z^2+x} \le \frac{1+y+yx}{(x+y+z)^2}

and


\frac{z}{z^3+x^2+y} \le \frac{1+z+zy}{(x+y+z)^2}.

Applying the AM-GM inequality


1 = \root{3}\of{xyz} \le \frac{x+y+z}3

and


xy+xz+yz \le xy+xz+yz + \frac{(x-y)^2+(x-y)^2+(y-z)^2}3 =
\frac13(x+y+z)^2,

we can establish


\frac{x}{x^3+y^2+z}+ \frac{y}{y^3+z^2+x}+ \frac{z}{z^3+x^2+y}\le


\le\frac{1+x+xz}{(x+y+z)^2}+\frac{1+y+yx}{(x+y+z)^2}+\frac{1+z+zy}{(x+y+z)^2}=
\frac{3\cdot1^2+(x+y+z)\cdot1+(xy+xz+yz)}{(x+y+z)^2} \le


\le
\frac{3\left(\frac{x+y+z}3\right)^2+
(x+y+z)\cdot\frac{x+y+z}3+\frac13(x+y+z)^2}{(x+y+z)^2}
= 1.


Statistics:

5 students sent a solution.
5 points:Backhausz Tibor, Frankl Nóra, Nagy 235 János, Nagy 648 Donát.
0 point:1 student.

Problems in Mathematics of KöMaL, September 2009