inequality involving $(xy+yz+zx)^3$ and the $pqr$ method

Question

For non-negative numbers $x$, $y$, and $z$ the claim is that:

$$(xy+yz+zx)^3+9 x^2 y^2 z^2 \geq 4 (x+y+z)(xy+yz+zx)xyz$$

Without loss of generality, one may assume that $x+y+z=1$ so that $xyz\le\frac{1}{27}$ and $xy+yz+zx\le\frac13$. I have not seen how Schur's inequality is helpful.

Answer 1

Let $xy=c$, $xz=b$ and $yz=a$.

Thus, we need to prove that $$(a+b+c)^3+9abc\geq4(a+b+c)(ab+ac+bc)$$ or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is Schur.