# Proving inequality $(a^2 + b^2)^3 \ge 32(a^3 + b^3)(ab - a - b)$

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

Since $$a + b ≥ 0$$, we can apply A.M.-G.M. inequality, I tried to apply the inequality, but wasn't able to reach a conclusive decision. How can I solve it using A.M.-G.M. inequality, or by any other way which is much easier than prior method.

• If $$ab\le a+b$$ your inequality becomes to be true. Commented Dec 18, 2019 at 16:01
• How did you reach to this conclusion? Commented Dec 18, 2019 at 16:49

We need to prove that: $$(a^2+b^2)^3+32(a+b)^2(a^2-ab+b^2)\geq32(a+b)(a^2-ab+b^2)ab$$ and since $$a+b\geq0,$$ it's enough to prove our inequality for $$ab\geq0,$$ which gives that $$a$$ and $$b$$ are non-negatives.
Now, by AM-GM $$(a^2+b^2)^3+32(a+b)^2(a^2-ab+b^2)\geq2\sqrt{(a^2+b^2)^3\cdot32(a+b)^2(a^2-ab+b^2)}.$$ Thus, it's enough to prove that: $$2\sqrt{(a^2+b^2)^3\cdot32(a+b)^2(a^2-ab+b^2)}\geq32(a+b)(a^2-ab+b^2)ab$$ or $$(a^2+b^2)^3\geq8(a^2-ab+b^2)a^2b^2.$$ Let $$a^2+b^2=2kab$$.
Thus, by AM-GM again $$k\geq1$$ and we need to prove that $$(2k)^3\geq8(2k-1)$$ or $$k^3-2k+1\geq0$$ or $$k^3-k^2+k^2-k-k+1\geq0$$ or $$(k-1)(k^2+k-1)\geq0,$$ which is true for $$k\geq1.$$
• How did you think of the manipulation that you did in the 1st step? And if $ab \leq 0$, in place of $\geq 0$, wouldn't the inequality always hold correct? Commented Dec 18, 2019 at 16:44
• If $ab\leq0$ so the inequality is obvious. Commented Dec 18, 2019 at 17:02