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Mathematical Olympiad in Ukraine, Third Round (January 23, ), 2010

Algebra Difficulty 5.0 AIME, harder Prove it Ukraine

Let $x$ , $y$ , $z$ be real numbers such that $x \in (0, 1]$ , $y \in (0, 1]$ , $z \in (0, 1]$ . Prove the inequality:
$\frac{x}{2+xy+yz} + \frac{y}{2+yz+zx} + \frac{z}{2+zx+xy} \le \frac{x+y+z}{x+y+z+xyz}$

Solution

Consider the obvious inequality:
$\begin{aligned} 0 \le (1-x)(1-y)(1-z) &= 1 + xy + yz + zx - x - y - z - xyz \\ &\Rightarrow 1 + xy + yz + zx \ge x + y + z + xyz \\ &\Rightarrow 2 + xy + yz \ge x + y + z + xyz \\ &\Rightarrow \frac{x}{2+xy+yz} \le \frac{x}{x+y+z+xyz}. \end{aligned}$
Analogously:
$\begin{aligned} \frac{y}{2+yz+zx} &\le \frac{y}{x+y+z+xyz}, \\ \frac{z}{2+zx+xy} &\le \frac{z}{x+y+z+xyz}. \end{aligned}$
Summing up yields the desired inequality.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.