Proving $a^4+b^4+c^4+(\sqrt {3}-1)(a^2 b c+a b^2 c+a b c^2 )\ge \sqrt {3} (a^3 b+b^3 c+c^3 a)$ for real $a$, $b$, $c$

Question

If $a$, $b$, $c$ are real numbers, I have to prove: $$a^4+b^4+c^4+(\sqrt {3}-1)(a^2 b c+a b^2 c+a b c^2 )\ge \sqrt {3} (a^3 b+b^3 c+c^3 a)$$

Since $$a^4+b^4+c^4 \ge abc(a+b+c)$$

then it is enough to prove

$$ abc(a+b+c) \ge a^3 b+b^3 c+c^3 a$$ or $$ a^2 b (c-a)+b^2 c (a-b)+a c^2 (b-c) \ge 0$$

I'm stuck there.

Answer 1

The hint.

Let $b=a+u$ and $c=a+v$.

Hence, we need to prove that $$(5-2\sqrt3)(u^2-uv+v^2)a^2+$$ $$+((4-\sqrt3)u^3-(2\sqrt3+1)u^2v+(\sqrt3-1)uv^2+(4-\sqrt3)v^3)a+$$ $$+u^4-\sqrt3u^3v+v^4\geq0,$$ which is a quadratic inequality of $a$ and it's enough to prove that $\Delta\leq0.$

I got $$\Delta=-(u^3+\sqrt3u^2v-(3+\sqrt3)uv^2+v^3)^2\leq0.$$

Answer 2

$$\Leftrightarrow \sum_{cyc}(\sqrt 3(a^2-b^2)-ab+2bc-ca)^2\ge 0$$