<s>[INST] Prove inequality $a^3b^3+2b^3c^3+3a^3c^3\le 0$ For $a^3+b^3+c^3=0$. Prove that $a^3b^3+2b^3c^3+3a^3c^3\le 0$

i think from $a^3+b^3+c^3=0$ we have one of the three numbers is zero, the other two are opposites.Suppose $a;b$ opposites and $c=0$
 [/INST] You can make:
$$a^3b^3+2b^3c^3+3a^3c^3=b^3(a^3+2c^3)+3a^3c^3$$
but, $a^3+2c^3=c^3-b^3$, so
$$b^3(c^3-b^3)+3a^3c^3=-b^6+c^3(b^3+3a^3)$$
but, $b^3+3a^3=-2b^3-3c^3$, so 
$$-b^6+c^3(-2b^3-3c^3)=-[(b^3+c^3)^2+2c^6]\le 0$$
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