Hello
is there any elegant way to prove the following,
$\displaystyle \sqrt{2a+1}+\sqrt{2b+1}\leq a+b+2,(a,b)\in \mathbb{R}^{+}$
Thanks
Dear Raoh,
Note that, $\displaystyle (a+1)^2=a^2+2a+1\geq{2a+1}~whenever~a\in\Re^{+}$
$\displaystyle \Rightarrow{a+1\geq\sqrt{2a+1}}~whenever~a\in\Re^+$
Similarly, $\displaystyle \Rightarrow{b+1\geq\sqrt{2b+1}}~whenever~a\in\Re^+$
Therefore, $\displaystyle \Rightarrow{a+1+b+1\geq\sqrt{2a+1}+\sqrt{2b+1}}~wh enever~a,b\in\Re^+$
$\displaystyle a+b+2\geq\sqrt{2a+1}+\sqrt{2b+1}~whenever~a,b\in\R e^+$
Hope this helps.