1. inequality [SOLVED]

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

2. 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.

3. i prefer this approach,
$\sqrt{2a+1}+\sqrt{2b+1}\leq \sqrt{2a+1+a^2}+\sqrt{2b+1+b^2}=a+b+2$