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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.

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

1 Answer 1

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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 $k\geq1.$

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  • $\begingroup$ 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? $\endgroup$
    – Zenix
    Commented Dec 18, 2019 at 16:44
  • $\begingroup$ If $ab\leq0$ so the inequality is obvious. $\endgroup$ Commented Dec 18, 2019 at 17:02

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