67th Czech and Slovak Mathematical Olympiad

If $x=y=z=t>0$ then the three fractions are sides of an equilateral triangle and $xy+yz+zx=3t^2$, hence $xy+yz+zx$ can attain all positive values. Similarly, for $x=y=t>0$ and $z=-2t$ the three fractions are $\frac{1}{3}{t}^{-2}$, $\frac{1}{3}{t}^{-2}$, $\frac{1}{6}{t}^{-2}$ which are positive numbers that are side-lengths of an isosceles triangle ($\frac{1}{6}<\frac{1}{3}+\frac{1}{3}$). Since $xy+yz+zx=-3t^2$, any negative value can be attained too.

Next we show that $xy+yz+zx$ can't be 0. Assume otherwise. Numbers $x,y,z$ are mutually distinct: if, say, $x$ and $y$ were equal then the denominator of the first fractions would be equal to $|x^2+2yz|=|xy+(yz+xz)|=0$ which is impossible.

Let's look at the fractions without absolute values. Subtracting $xy+yz+zx=0$ from each denominator we get $$\begin{array}{rl}\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}& =\\ & =\frac{1}{(x-y)(x-z)}+\frac{1}{(y-z)(y-x)}+\frac{1}{(z-x)(z-y)}=\\ & =\frac{(z-y)+(x-z)+(y-x)}{(x-y)(y-z)(z-x)}=0.\end{array}$$

This implies that among the original fractions (with absolute values), one of them is a sum of the other two. Hence the fractions don't fulfil triangle inequality and we reached the desired contradiction.