Prove : $2(x^3+y^3+z^3)+3xyz \ge 3(x^2y+y^2z+z^2x)$ with $x,y,z \gt 0$ At first I tried to divide both side by $xyz$, the inequality became:
$$2\sum {\frac{x^2}{yz}}+3 \ge 3\sum{\frac xy}$$
Let $$\frac xy = a;\frac yz = b;\frac zx = c;$$
So all we have to prove is
$$2\sum \frac ab +3 \ge 3 \sum a $$ with $a,b,c$ being positive real and $abc=1$ .
And then I stuck at this point. Any help ?

By Schur inequality (whenever you get $xyz$ with positive coefficient on the higher side, worth trying Schur):
$$x^3+y^3+z^3 + 3xyz \geqslant (x^2y+y^2z+z^2x) + (xy^2+yz^2+zx^2)$$
So it is enough to show
$$x^3+y^3+z^3+(xy^2+yz^2+zx^2) \geqslant 2(x^2y+y^2z+z^2x)$$
which follows from three AM-GMs like $x^3+xy^2 \geqslant 2x^2y$.
Equality is iff $x=y=z$.
