# Symmetric homogeneous fourth-degree inequality in three variables

Does anyone know how to show that for any $a,b,c>0$,

$$(a^3+b^3+c^3+9abc)(a+b+c) \geq 4(ab+ac+bc)^2$$

I could only think of Newton’s inequality, which unfortunately goes the other way : it shows that $abc\ \times \frac{a+b+c}{3} \leq \left( \frac{ab+ac+bc}{3}\right)^2$.

By third degree Schur's inequality, we have $$a^3+b^3+c^3+3abc \ge \sum_{cyc} ab(a+b)$$ So it is enough to show that $$\left(\sum_{cyc} ab(a+b)+6abc\right)(a+b+c) \ge 4(ab+bc+ca)^2$$ which is equivalent on expansion (ugh) to the obvious $$\sum_{cyc}ab(a-b)^2 \ge 0$$
P.S. Equality will be like in Schur, i.e. when $a=b=c$ (or when two of the variables are equal and the other zero, if that's allowed).