
2. Let $x, y, z$ be real numbers such that

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

are side-lengths of a (non-degenerate) triangle. Find all possible values of $x y+$ $y z+z x$.

(Michal Rolínek)


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

Next we show that $x y+y z+z x$ 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 $\left|x^{2}+2 y z\right|=|x y+(y z+x z)|=0$ which is impossible.

Let's look at the fractions without absolute values. Subtracting $x y+y z+z x=0$ from each denominator we get

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

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.

Answer. Possible values are all real numbers except for 0 .
