Proving the inequality $(a^2-ab+b^2)(c^2-ac+a^2)(b^2-bc+c^2) \le 12$. This is a follow up question to my previous post "Inequalities of expressions completely symmetric in their variables". An answer provided a counterexample to me reasoning: under the constraints $a,b,c\in\Bbb{R}^+$ and $a+b+c=3$,
$$
(a^2-ab+b^2)(c^2-ac+a^2)(b^2-bc+c^2) \le 12.
$$
I demanded a proof for this inequality, however since it was an entirely different question, I felt the need for a new post.

Let $a\geq b\geq c\geq0$.
Hence,
$$\prod_{cyc}(a^2-ab+b^2)\leq(a^2-ab+b^2)a^2b^2=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2=$$
$$=12(3-ab)\cdot\frac{ab}{2}\cdot\frac{ab}{2}\leq12\left(\frac{3-ab+\frac{ab}{2}+\frac{ab}{2}}{3}\right)^3=12$$
