Real numbers $x,y,z$ are selected so, that $\frac{1}{|x^2 +2yz|}$, $\frac{1}{|y^2 +2xz|}$, $\frac{1}{|z^2 +2xy|}$ are sides of a triangle(they satisfy the triangle inequality). Determine the possible values of the expression $xy+xz+yz$.
It is easy to prove that all positive and negative real numbers can be expressed. What about $0$? Any help is appreciated.
Suppose that $xy+yz+zx=0$. Then, we have the following identities $$ \frac{1}{|x^2+2yz|}=\frac{1}{|x^2+2yz-xy-zx-yz|}=\frac{1}{|x-y|\cdot |x-z|}, \\ \frac{1}{|y^2+2zx|}=\frac{1}{|y^2+2zx-xy-zx-yz|}=\frac{1}{|x-y|\cdot |y-z|}, \\ \frac{1}{|x^2+2yz|}=\frac{1}{|z^2+2xy-xy-zx-yz|}=\frac{1}{|z-x|\cdot |y-z|}. $$ Now, it's easy to prove that sum of two of these number is equal to the third number, so they doesn't satisfy the triangle inequality. Hence, $xy+yz+zx\neq 0$, as desired.