, 2008

Solution:
The idea is to split the inequality in two, showing that
$$\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}$$
can be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has
$$(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq \left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}$$
On the other hand, as
$$\sqrt{\frac{x}{y}} \geq \frac{2x}{x y+1} \Leftrightarrow (\sqrt{x y}-1)^{2} \geq 0$$
by summation one has
$$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}} \geq \frac{2x}{x y+1}+\frac{2y}{y z+1}+\frac{2z}{z x+1}$$
The rest is obvious.