Since $xyz \neq 0$, we can do the following transformation of the given expression:
$$\begin{aligned} \frac{1}{x^2+2yz} + \frac{1}{y^2+2zx} + \frac{1}{z^2+2xy} &= \frac{1}{x^2+2yz-xy-yz-zx} + \frac{1}{y^2+2zx-xy-yz-zx} + \frac{1}{z^2+2xy-xy-yz-zx} \\ &= \frac{1}{x^2+yz-xy-zx} + \frac{1}{y^2+zx-xy-yz} + \frac{1}{z^2+xy-yz-zx} \\ &= \frac{1}{(x-y)(x-z)} + \frac{1}{(y-x)(y-z)} + \frac{1}{(z-x)(z-y)} \\ &= -\frac{1}{(x-y)(z-x)} - \frac{1}{(x-y)(y-z)} - \frac{1}{(z-x)(y-z)} \\ &= -\frac{(y-z)+(z-x)+(x-y)}{(x-y)(z-x)(y-z)} = 0. \end{aligned}$$

Thus, there are no zeros among the three terms on the left-hand side, and it turned out that the sum of two is equal to the third modulo, that is, they can not be the sides of a non-degenerate triangle. The resulting contradiction completes the proof.