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
$$\begin{aligned} \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy} &= \\ &= \frac{1}{(x-y)(x-z)} + \frac{1}{(y-z)(y-x)} + \frac{1}{(z-x)(z-y)} = \\ &= \frac{(z-y) + (x-z) + (y-x)}{(x-y)(y-z)(z-x)} = 0. \end{aligned}$$

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.