<s>[INST] If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$ 
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$

This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be able to cancel anything out. I am also struggling with other methods like AM-GM since $a,b$ are not necessarily positive. Any help would be appreciated.
 [/INST] We need to prove that
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq32(a^3+b^3)ab,$$ which is obviously true for $ab<0$.
Let $ab\geq0.$ 
Thus, by AM-GM three times we obtain:
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq2\sqrt{(a^2+b^2)^3\cdot32(a^3+b^3)(a+b)}=$$
$$=8\sqrt{2(a^2-ab+b^2+ab)^3(a^3+b^3)(a+b)}\geq8\sqrt{2\left(2\sqrt{(a^2-ab+b^2)\cdot ab}\right)^3(a^3+b^3)(a+b)}=$$
$$=32\sqrt{(a^3+b^3)^2\sqrt{(a^2-ab+b^2)a^3b^3}}\geq32\sqrt{(a^3+b^3)^2\sqrt{ab\cdot a^3b^3}}=32(a^3+b^3)ab.$$
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