Serbian Mathematical Olympiad

Solution:

Let us denote $a=\frac{x-y}{x y+2 y+1}$, $b=\frac{y-z}{y z+2 z+1}$ and $c=\frac{z-x}{z x+2 x+1}$. Then $1+\frac{1}{a}=\frac{x y+x+y+1}{x-y}$ and from this $\frac{a}{a+1}=\frac{x-y}{x y+x+y+1}=\frac{1}{y+1}-\frac{1}{x+1}$; similarly $\frac{b}{b+1}=\frac{1}{z+1}-\frac{1}{y+1}$ and $\frac{c}{c+1}=\frac{1}{x+1}-\frac{1}{z+1}$.
From $0<\frac{1}{x+1}, \frac{1}{y+1}, \frac{1}{z+1}<1$ it follows that $\frac{a}{a+1}, \frac{b}{b+1}, \frac{c}{c+1}<1$, so $a+1, b+1, c+1$ are positive. Moreover, we have $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=0$, i.e. $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=3$. Now by the Cauchy-Schwarz inequality $(a+1)+(b+1)+(c+1) \geqslant 3$, i.e. $a+b+c \geqslant 0$.

Second solution. Multiplying out and grouping, the required inequality reduces to
$$\begin{aligned} & 2(x-1)^{2}(y-z)^{2}+2(y-1)^{2}(z-x)^{2}+2(z-1)^{2}(x-y)^{2} \\ & +9\left(x y^{2}+y z^{2}+z x^{2}-3 x y z\right)+3\left(x^{2} y+y^{2} z+z^{2} x-3 x y z\right) \geqslant 0 \end{aligned}$$
where all the terms are nonnegative by the AM-GM inequality.