Show $(a^2 + b^2 + c^2 + d^2)^3 \ge 3 (a^3 + b^3 + c^3 + d^3)^2$ Let $a,b,c,d$ be real numbers such that  $a + b + c +d  = 0$.
Show:
$(a^2 + b^2 + c^2 + d^2)^3 \ge 3 (a^3 + b^3 + c^3 + d^3)^2.$
I've applied AM-GM to the left hand side to find  $ (a^2 + b^2 + c^2 + d^2) \ge  4 \cdot\sqrt[4]{a^2b^2c^2d^2}, $ so $  (a^2 + b^2 + c^2 + d^2)^3 \ge  64 \cdot (abcd)^{3/2}.$
And I've been trying to get the right hand side bounded by a lower value but haven't a way forward.

We need to prove that $$(a^2+b^2+c^2+(a+b+c)^2)^3\geq3(a^3+b^3+c^3-(a+b+c)^3)^2$$ or
$$\left(\sum_{cyc}(a+b)^2\right)^3\geq27(a+b)^2(a+c)^2(b+c)^2,$$ which is just AM-GM:
$$\frac{\sum\limits_{cyc}(a+b)^2}{3}\geq\sqrt[3]{\prod_{cyc}(a+b)^2}.$$
