An algebraic inequality

Question

Let $a,b,c$ be real numbers with $a+b+c=0$. I want to show that $$a^2b^2+b^2c^2+c^2a^2+3 \geq 6abc$$

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Since, $a^2b^2+a^2c^2+b^2c^2=9v^4-6uw^3$, we see that our inequality is a linear inequality of $w^3$,

which says that it's enough to prove our inequality for an extremal value of $w^3$,

which happens for equality case of two variables.

Let $b=a$ and $c=-2a$.

Hence, we need to prove that: $$3a^4-4a^3+1\geq0,$$ which is AM-GM: $$3a^4+1\geq4\sqrt[4]{(a^4)^3\cdot1}=4|a^3|\geq4a^3.$$ Done!

Another way.

Since $a+b+c=0$, we can assume $c\leq0$. Let $ab=x$.

Thus, $a^2+b^2=(a+b)^2-2ab=c^2-2x$ and we need to prove that $$x^2-2(c^2+3c)x+c^4+3\geq0$$ for which it's enough to prove that $$(c^4+3c)^2-(c^4+3)\leq0$$ or $$(c+1)^2(2c-1)\leq0.$$ Done!