59th Ukrainian National Mathematical Olympiad

Since $xyz\ne 0$, we can do the following transformation of the given expression: $$\begin{array}{rl}\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}& =\frac{1}{x^2+2yz-xy-yz-zx}+\frac{1}{y^2+2zx-xy-yz-zx}+\frac{1}{z^2+2xy-xy-yz-zx}\\ & =\frac{1}{x^2+yz-xy-zx}+\frac{1}{y^2+zx-xy-yz}+\frac{1}{z^2+xy-yz-zx}\\ & =\frac{1}{(x-y)(x-z)}+\frac{1}{(y-x)(y-z)}+\frac{1}{(z-x)(z-y)}\\ & =-\frac{1}{(x-y)(z-x)}-\frac{1}{(x-y)(y-z)}-\frac{1}{(z-x)(y-z)}\\ & =-\frac{(y-z)+(z-x)+(x-y)}{(x-y)(z-x)(y-z)}=0.\end{array}$$

Thus, there are no zeros among the three terms on the left-hand side, and it turned out that the sum of two is equal to the third modulo, that is, they can not be the sides of a non-degenerate triangle. The resulting contradiction completes the proof.