Find the largest real number $\alpha$ such that, for all non-negative real numbers $x, y$ and $z$, the following inequality holds:

$$
(x+y+z)^{3}+\alpha\left(x^{2} z+y^{2} x+z^{2} y\right) \geq \alpha\left(x^{2} y+y^{2} z+z^{2} x\right) .
$$

Answer: $6 \sqrt{3}$.
Without loss of generality, $x$ is the largest amongst the three variables. By moving $\alpha\left(x^{2} z+y^{2} x+z^{2} y\right)$ to the right-hand side and factoring, we get the equivalent inequality

$$
(x+y+z)^{3} \geq \alpha(x-y)(x-z)(y-z)
$$

If $z>y$, then the right-hand side is non-positive, so we can assume $x \geq y \geq z$.
Note that

$$
\begin{aligned}
x+y+z & \geq x+y-2 z \\
& =\frac{1}{\sqrt{3}}(x-y)+\left(1-\frac{1}{\sqrt{3}}\right)(x-z)+\left(1+\frac{1}{\sqrt{3}}\right)(y-z) \\
& \geq 3 \sqrt[3]{\frac{2}{3 \sqrt{3}}(x-y)(x-z)(y-z)} .
\end{aligned}
$$

Cubing both sides gives $(x+y+z)^{3} \geq 6 \sqrt{3}(x-y)(x-z)(y-z)$. The equality holds when $z=0$ and $x=y(2+\sqrt{3})$. So $\alpha=6 \sqrt{3}$.