Austria competition problems, 2014

We first note that the left side of the inequality is certainly non-negative for all values of $x$, $y$ and $z$. If any of the variables is equal to $0$, the right-hand side is equal to $0$, and the inequality certainly holds. If equality holds with any variable being equal to $0$, we can without loss of generality consider the case where $x = 0$. In this case, the inequality reduces to $y^4 z^4 \geq 0$, and equality holds if either $y = 0$ or $z = 0$. We note that all triples $(0, 0, t)$, $(0, t, 0)$ and $(t, 0, 0)$ yield equality for any integer values of $t$.

We can now consider the case in which no variable is equal to $0$. In this case, the AM-GM inequality gives us
$$\begin{aligned} (x^2 + y^2 z^2) \cdot (y^2 + x^2 z^2) \cdot (z^2 + x^2 y^2) &\geq 8 \cdot (\sqrt{x^2 y^2 z^2})^3 \\ &= 8 \cdot |x|^3 |y|^3 |z|^3, \end{aligned}$$
and since $|x|^3 \geq |x| \geq x$, $|y|^3 \geq |y|^2 = y^2$ and $|z|^3 \geq z^3$ hold for any integer values of $x$ and $y$, the proof is complete. Equality holds for $x^2 = y^2 = z^2 = 1$ and $|x||z|^3 = xz^3$, i.e. for $|x| = |y| = |z| = 1$ if $x$ and $z$ have the same sign. We see that further cases of equality are given by $(1, 1, 1)$, $(1, -1, 1)$, $(-1, 1, -1)$ and $(-1, -1, -1)$.