Team Selection Test

Since $\sqrt{x}$, $\sqrt{y}$, $\sqrt{z}$ are sides of a triangle, we have $$(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{y}+\sqrt{z}-\sqrt{x})(\sqrt{z}+\sqrt{x}-\sqrt{y})=2(xy+yz+zx)-x^2-y^2-z^2\ge 0.$$ We also have $$5xy=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)xy=x^2+\frac{x{y}^2}{z}+yz$$ $$5yz=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)yz=y^2+\frac{y{z}^2}{x}+zx$$ $$5zx=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)zx=z^2+\frac{z{x}^2}{y}+xy$$ and by summing up the equations above, we get $$2(xy+yz+zx)-(x^2+y^2+z^2)=\frac{x{y}^2}{z}+\frac{y{z}^2}{x}+\frac{z{x}^2}{y}-2(xy+yz+zx)$$ $$=\frac{x(y^2-2z^2)}{z}+\frac{y(z^2-2x^2)}{x}+\frac{z(x^2-2y^2)}{y}\ge 0.$$