50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

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