parmenides51 wrote: Let the real numbers $x, y, z$ be such that $x + y + z = 0$. Prove that $$6(x^3 + y^3 + z^3)^2 \le (x^2 + y^2 + z^2)^3.$$ https://artofproblemsolving.com/community/c6h1387404p7720007 https://artofproblemsolving.com/community/c6h1643102p10358848 https://artofproblemsolving.com/community/c4h1080080p4740681 https://math.stackexchange.com/questions/3817541/proving-6x3y3z32-leq-x2y2z23-where-xyz-0?noredirect=1

Since $x+y+z=0$, we have $x^3+y^3+z^3 = 3xyz$ and $x^2+y^2+z^2=-2(xy+xz+yz)$, so we must show $6(3xyz)^2 \leq (-2(xy+xz+yz))^3$, or $6(3xyz)^2 \leq -8(xy+xz+yz)^3$. However how do I show this??