Mathematica competitions in Croatia

The given inequality can be written in the following form: $$\alpha (x^3+y^3+z^3-x^{2}y-y^{2}z-z^{2}x)\ge x^{2}y+y^{2}z+z^{2}x-x{y}^2-y{z}^2-z{x}^2.$$ From the rearrangement inequality it follows that the left-hand side of the previous inequality is positive. Therefore, it suffices to prove the desired inequality for $\alpha =\frac{1}{2}$: $$x^3+y^3+z^3+2(x{y}^2+y{z}^2+z{x}^2)\ge 3(x^{2}y+y^{2}z+z^{2}x).(1)$$ Since the given inequality is cyclic, without loss of generality we may assume that $z\le x$ and $z\le y$. Let $a,b\ge 0$ be real numbers such that $x=z+a$, $y=z+b$. Therefore, the inequality (1) can be written as follows: $$a^3-3a^{2}b+2a{b}^2+b^3+2z(a^2-ab+b^2)\ge 0.$$ By AM-GM inequality, we have: $$a^2-ab+b^2\ge ab\ge 0.$$ Thus, it suffices to show: $$a^3+2a{b}^2+b^3-3a^{2}b\ge 0.$$ However, this can be written as follows: $$\begin{array}{rl}& a^3-4a^{2}b+4a{b}^2+a^{2}b-2a{b}^2+b^3\ge 0,\\ & a(a^2-4ab+4b^2)+b(a^2-2ab+b^2)\ge 0,\\ & a(a-2b{)}^2+b(a-b{)}^2\ge 0,\end{array}$$ which obviously holds.