inequality

**Raoh** · November 25th 2010, 02:44 PM

Hello
is there any elegant way to prove the following,
$\sqrt{2a+1}+\sqrt{2b+1}\leq a+b+2,(a,b)\in \mathbb{R}^{+}$
Thanks

**Sudharaka** · November 26th 2010, 05:53 AM

Originally Posted by Raoh

Hello
is there any elegant way to prove the following,
$\sqrt{2a+1}+\sqrt{2b+1}\leq a+b+2,(a,b)\in \mathbb{R}^{+}$
Thanks

Dear Raoh,

Note that, $(a+1)^2=a^2+2a+1\geq{2a+1}~whenever~a\in\Re^{+}$

$\Rightarrow{a+1\geq\sqrt{2a+1}}~whenever~a\in\Re^+$

Similarly, $\Rightarrow{b+1\geq\sqrt{2b+1}}~whenever~a\in\Re^+$

Therefore, $\Rightarrow{a+1+b+1\geq\sqrt{2a+1}+\sqrt{2b+1}}~wh enever~a,b\in\Re^+$

$a+b+2\geq\sqrt{2a+1}+\sqrt{2b+1}~whenever~a,b\in\R e^+$

Hope this helps.

**Raoh** · November 26th 2010, 10:30 AM

i prefer this approach,
$\sqrt{2a+1}+\sqrt{2b+1}\leq \sqrt{2a+1+a^2}+\sqrt{2b+1+b^2}=a+b+2$
Your solution was awesome.
Thank you.

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