Estonian Math Competitions

If $x = y = z = \frac{2012}{3}$, then
$$\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)} = 2012.$$
Now we prove that for all $x$, $y$, $z$ satisfying the premises we have
$$\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)} \le 2012$$
It suffices to show that $(x^2 + y^2 + z^2)(x^3 + y^3 + z^3) \le 2012(x^4 + y^4 + z^4)$, or $(x^2 + y^2 + z^2)(x^3 + y^3 + z^3) \le (x + y + z)(x^4 + y^4 + z^4)$. Multiplying out, simplifying and rearranging the terms gives $xy(x - y)(x^2 - y^2) + xz(x - z)(x^2 - z^2) + yz(y - z)(y^2 - z^2) \ge 0$. Since the differences in the brackets in every product have equal signs, the products are non-negative, showing that the necessary inequality holds.