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

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 $$a^6+a^4b^2+a^4b^2+a^4b^2+16a^4+16a^4+16a^3b+16a^3b \geq 32a^4b.$$

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 $\ge 0$.

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.