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.

Equivalently, you want that 
$$
f = \frac18((a^2+b^2)^3 - 32(a^3+b^3)(ab-a-b)) \ge 0
$$
Let $a = x+y$ and $b = x-y$. Then we have $x \ge 0$ and arbitrary $y$. 
Putting in these variables you have that 
$$
f = (x - 4)^2 x^4 + y^2 (3 ((x- 8/3)^2 +  80/9) x^2 + y^2 (x (3 x + 24) + y^2))
$$
which is clearly $\ge 0$.
