2. Real numbers $x, y, z$ are chosen such that the numbers

$$
\frac{1}{\left|x^{2}+2 y z\right|}, \quad \frac{1}{\left|y^{2}+2 z x\right|}, \quad \frac{1}{\left|z^{2}+2 x y\right|}
$$

are the lengths of the sides of a (non-degenerate) triangle. Determine all possible values of the expression $x y+y z+z x$.

(Michal Rolínek)
Solution. When choosing $x=y=z=t>0$, the mentioned numbers are the lengths of the sides of an equilateral triangle, and $x y+y z+z x=3 t^{2}$, so the expression $x y+y z+z x$ can take all positive values. Similarly, for $x=y=t>0$ and $z=-2 t$, the three fractions have values $\frac{1}{3} t^{-2}$, $\frac{1}{3} t^{-2}$, and $\frac{1}{6} t^{-2}$, which are positive numbers corresponding to the lengths of the sides of an isosceles triangle (the inequality $\frac{1}{6}<\frac{1}{3}+\frac{1}{3}$ holds). In this case, $x y+y z+z x=-3 t^{2}$, so the expression $x y+y z+z x$ can also take all negative values. [^0]

We further prove that the expression $x y+y z+z x$ cannot take the value zero. Assume the opposite. The numbers $x, y, z$ must be pairwise distinct: if, for example, $x=y$, the denominator of the first fraction would be $\left|x^{2}+2 y z\right|=|x y+(y z+x z)|=0$, which is impossible.

Consider the fractions without absolute values. By subtracting $x y+y z+z x=0$ from each denominator and then simplifying to a product, we get

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

This implies that in the original triplet of fractions (with absolute values), the value of one of them was the sum of the values of the other two. This contradicts the assumption that these values are the lengths of the sides of a non-degenerate triangle (they cannot satisfy the triangle inequality!).

Answer. The possible values of the expression are all real numbers except 0.