# Inequalities for the term $(a+b)^2$ [closed]

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?

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

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

$$(a+b)^2 \ge ka^2+lb^2$$ where $k,l \le 1$ and $ab \ge 0$.
• Is it $a,b\geq 0$? – zorro47 Oct 3 '17 at 20:42
• 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. – zorro47 Oct 3 '17 at 20:45
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).