Let $x$, $y$, $z$ be positive. Prove an inequality... Let $x$, $y$, $z$ be positive. Prove the inequality
$$4(x+y+z)^3 \ge 27(yx^2+zy^2+xz^2+xyz)$$

I have no idea of how the proof should look like, tried to get rid of the braces but it seems to be the wrong way, as the equation becomes very long and has very different positive and negative coefficients, so doing a factorization is quite impossible.

I guess some kind of theorem or something can be applied here, but have no idea which one
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x+xyz=xy\cdot x+yz\cdot y+zx\cdot z+xyz\leq ab\cdot a+ac\cdot b+bc\cdot c+abc=$$
$$=b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}(x+y+z)^3$$
Done!
Let z is parameter, x+y = t is parameter.
Try to proof what right side is maximal if x = y.
And simple  Weierstrass  works