How to prove inequality $(a^2+b^2+c^2)^3\ge6(a^3+b^3+c^3)^2$ when $a+b+c=0$? I know the identity $a^3+b^3+c^3-3abc = 1/2 (a+b+c) [(a-b)^2+(b-c)^2+(c-a)^2]$.
So when it comes to this problem where $a+b+c=0$, you get $(a^2+b^2+c^2)^3\ge54(abc)^2$ when $a+b+c=0$
(because $ a^3+b^3+c^3-3abc=0$).
But if you use average inequality, you can only get $(a^2+b^2+c^2)^3\ge27(abc)^2$.
So what kind of technique works on this problem?

After substitution $c=-a-b$ we need to prove that
$$4(a^2+ab+b^2)^3\geq27a^2b^2(a+b)^2,$$ which is true because
$$4(a^2+ab+b^2)^3-27a^2b^2(a+b)^2=(a-b)^2(2a+b)^2(a+2b)^2\geq0.$$
Indeed. 
let $a^2+b^2=2uab$. 
Thus,
$$4(a^2+ab+b^2)^3-27a^2b^2(a+b)^2=4(2uab+ab)^3-27a^2b^2(2uab+2ab)=$$
$$=2a^3b^3(2(2u+1)^3-27(u+1))=2a^3b^3(16u^3+24u^2-15u-25)=$$
$$=2a^3b^3(16u^3-16u^2+40u^2-40u+25u-25)=2a^3b^3(u-1)(16u^2+40u+25)=$$
$$=2a^3b^3(u-1)(4u+5)^2=(a^2+b^2-2ab)(2(a^2+b^2)+5ab)^2=(a-b)^2(2a+b)^2(b+2a)^2.$$
There is a nice proof by AM-GM.
Indeed, since $a+b+c=0$, we can assume that $ab\leq0.$
Thus, by AM-GM we obtain:
$$(a^2+b^2+c^2)^3=(a^2+b^2+(a+b)^2)^3=(2(a+b)^2-ab-ab)^3\geq$$
$$\geq\left(3\sqrt[3]{2(a+b)^2\cdot(-ab)^2}\right)^3=54a^2b^2(a+b)^2=$$
$$=6(3ab(a+b))^2=6(a^3+b^3-(a+b)^3)^2=6(a^3+b^3+c^3)^2.$$
