Balti Tee 2024 · Ülesanne 4

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 $\alpha=6 \sqrt{3}$.