Problem 1294

Proof: Assume $x y z=1$.
First, prove $\quad \frac{x z}{x^{2}+x z+y z} \leqslant \frac{1}{1+y+\frac{1}{z}}$.
(1) is equivalent to
$$x^{2}+x z+y z \geqslant x z+1+x$$

which is $x^{2}+\frac{1}{x} \geqslant 1+x$, or $(x-1)^{2}(x+1) \geqslant 0$. Hence (1) holds.
Also, since
$$\frac{1}{1+y+\frac{1}{z}}=\frac{z}{y z+z+1}$$

and similarly,
$$\begin{array}{l} \text { we have } \frac{x y}{y^{2}+x y+x z} \leqslant \frac{1}{1+z+\frac{1}{x}}=\frac{1}{1+z+y z} \\ \frac{y z}{z^{2}+y z+x y} \leqslant \frac{1}{1+x+\frac{1}{y}}=\frac{x y z}{x y z+x+x z}=\frac{y z}{1+z+y z} \end{array}$$

thus $\sum_{\mathrm{c} x} \frac{x z}{x^{2}+x z+y z} \leqslant \frac{1+z+y z}{1+z+y z}=1$, the conclusion is established.