Tintarn wrote: Find the largest real number $\alpha$ such that, for all non-negative real numbers $x$, $y$ and $z$, the following inequality holds: \[ (x+y+z)^3 + \alpha (x^2z + y^2x + z^2y) \geq \alpha (x^2y + y^2z + z^2x). \] $6\sqrt3$ of course.

Here is how to get it. Note that this is not a full solution but one can be put together (possibly with certain steps reversed) based on what is given below. When looking for the largest value it is reasonable to try $\alpha >0$ first. The inequality is equivalent to $$(x+y+z)^3\ge\alpha (x-y)(x-z)(y-z)$$so it suffices to consider $x\ge y\ge z$. Take $1>\beta>0$. We have by AM-GM $$\begin{aligned}(x+y+z)^3&=\left(\beta (x-y)+(1-\beta )(x-z)+(1+\beta )(y-z)+3z\right)^3\\&\ge \left(\beta (x-y)+(1-\beta )(x-z)+(1+\beta )(y-z)\right)^3\\&\ge 27\beta\left(1-\beta ^2\right)(x-y)(x-z)(y-z)\end{aligned}$$We want to find the largest value that $27\beta\left(1-\beta ^2\right)$ can attain. Again by AM-GM $$\left(27\beta\left(1-\beta ^2\right)\right)^2=\frac{27^2}{2}\cdot 2\beta ^2\cdot\left(1-\beta ^2\right)^2\le\frac{27^2}{2}\left(\frac{2\beta ^2+2\left(1-\beta ^2\right)}{3}\right)^3=\left(6\sqrt{3}\right)^2$$Clearly, there exists a $\beta$ such that $2\beta ^2=1-\beta ^2$, namely $\beta =\frac{1}{\sqrt{3}}$. For equality to hold we must then have $z=0$ and $\frac{1}{\sqrt{3}}(x-y)=\left(1-\frac{1}{\sqrt{3}}\right)x=\left(1+\frac{1}{\sqrt{3}}\right)y$.

Tintarn wrote: Find the largest real number $\alpha$ such that, for all non-negative real numbers $x$, $y$ and $z$, the following inequality holds: \[(x+y+z)^3 + \alpha (x^2z + y^2x + z^2y) \geq \alpha (x^2y + y^2z + z^2x).\] It's old: https://artofproblemsolving.com/community/c6h1144659p5395326