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Do you know any inequalities "$\ge$" for $(a+b)^2$? I proved a very simple one

$$(a+b)^2\geq \frac{1}{2}a^2-b^2.$$

Do you know any other inequalities?

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closed as unclear what you're asking by José Carlos Santos, user223391, Xander Henderson, user99914, Leucippus Oct 4 '17 at 2:40

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  • $\begingroup$ What makes an inequality better than another one? $\endgroup$ – user228113 Oct 3 '17 at 20:19
  • $\begingroup$ or do you mean $$(a+b)^2 \geq \frac{1}{2}(a^2-b^2)$$? $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '17 at 20:22
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How about, in general:

$$(a+b)^2 \ge ka^2+lb^2$$ where $k,l \le 1$ and $ab \ge 0$.

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  • $\begingroup$ Is your inequality really true? It looks very spectacular. I think it is - that's what I've been looking for! $\endgroup$ – zorro47 Oct 3 '17 at 20:34
  • $\begingroup$ @Vessemir I had missed 1 condition. Added it. Its true. Try to prove it. $\endgroup$ – SchrodingersCat Oct 3 '17 at 20:41
  • $\begingroup$ Is it $a,b\geq 0$? $\endgroup$ – zorro47 Oct 3 '17 at 20:42
  • $\begingroup$ @Vessemir No. The product must be non negative. Else it wont hold. $\endgroup$ – SchrodingersCat Oct 3 '17 at 20:43
  • $\begingroup$ Defenitely, but $a,b\geq 0$ implpies that and it's good enough for me, since in my case I want $a$ and $b$ to norms in a Banach space. $\endgroup$ – zorro47 Oct 3 '17 at 20:45
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By the AM-GM inequality, if $a,b$ are nonnegative, we have $$\frac{a+b}{2}\ge\sqrt{ab},$$ whence $$(a+b)^2\ge 4ab.$$ This inequality is optimal in some sense, because the equality holds iff $a=b$. So, it could not be refined.

If $ab<0$ then of course the inequality holds (it is not optimal in this case). If both $a,b$ are negative, then $-a,-b$ are positive and the inequality also holds true (optimal).

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