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.
 A: Assuming that $a,b > 0$, you want to prove that:
$$a^6+3a^4b^2+3a^2b^4+b^6+32a^4+32a^3b+32ab^3+32b^4 \geq 32 a^4b+ 32ab^4.$$
Now, show that $$a^6+a^4b^2+a^4b^2+a^4b^2+16a^4+16a^4+16a^3b+16a^3b \geq 32a^4b.$$
A: 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 $\ge 0$.
A: 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.
A: 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:
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq2\sqrt{(a^2+b^2)^3\cdot32(a^3+b^3)(a+b)}=$$
$$=8\sqrt{2(a^2-ab+b^2+ab)^3(a^3+b^3)(a+b)}\geq8\sqrt{2\left(2\sqrt{(a^2-ab+b^2)\cdot ab}\right)^3(a^3+b^3)(a+b)}=$$
$$=32\sqrt{(a^3+b^3)^2\sqrt{(a^2-ab+b^2)a^3b^3}}\geq32\sqrt{(a^3+b^3)^2\sqrt{ab\cdot a^3b^3}}=32(a^3+b^3)ab.$$
