Mathematica competitions in Croatia

The A-H inequality together with the condition of the problem gives $$xy+yz+zx=xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge xyz\cdot \frac{9}{x+y+z}=\frac{9xyz}{18xyz}=\frac{1}{2}.$$ Using $1\le 2xy+2yz+2zx$ we get $$x^2+2yz+1\le x^2+2xy+2zx+4yz=(x+2y)(x+2z)\le (x+y+z{)}^2$$ where the last inequality is a consequence of the A-G inequality. Hence $\frac{x}{\sqrt{x^2+2yz+1}}\ge \frac{x}{x+y+z}$ and analogously $\frac{y}{\sqrt{y^2+2zx+1}}\ge \frac{y}{x+y+z}$, $\frac{z}{\sqrt{z^2+2xy+1}}\ge \frac{z}{x+y+z}$. Adding these three inequalities finishes the proof.