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.

With variable subsitutions $a = x + y$, $b = x - y$, the inequality becomes  $$(x^2 + y^2)^3 - 8 (x^3 + xy^2)(x^2 - y^2 - 2x) \geq 0$$ in the domain $x \geq 0$. Denote the LHS $f(x, y)$. As $f(x, y) = f(x, -y)$, it suffices to prove $f \geq 0$ in the closed first quadrant.
First note that $$f(x, 0) = x^6 - 8x^5 + 16 x^4 = x^4 (x-4)^2 \geq 0$$ so the inequality holds on the $x$-axis. Routine computation gives $$ \frac{\partial f}{\partial y} = 6x^4 y + 12x^2 y^3 + 6y^5 +  32 xy^3 + 32 x^2 y$$ which is clearly nonnegative in the first quadrant, so we have $f(x, y) \geq f(x, 0) \geq 0$ for all $x \in \mathbb{R}_{\geq 0}, y \in \mathbb{R}$, and we're done.
