Austria 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 $y^4{z}^4\ge 0$, 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 $$\begin{array}{rl}(x^2+y^2{z}^2)\cdot (y^2+x^2{z}^2)\cdot (z^2+x^2{y}^2)& \ge 8\cdot (\sqrt{x^2{y}^2{z}^2}{)}^3\\ & =8\cdot |x{|}^3|y{|}^3|z{|}^3,\end{array}$$ and since $|x{|}^3\ge |x|\ge x$, $|y{|}^3\ge |y{|}^2=y^2$ and $|z{|}^3\ge z^3$ hold for any integer values of $x$ and $y$, the proof is complete. Equality holds for $x^2=y^2=z^2=1$ and $|x||z{|}^3=x{z}^3$, 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 $(-1,-1,-1)$.