What would you suggest for the following inequality? $$\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$$
Thanks in advance!
Sis.
EDIT: Based upon the nice solution provided by Sasha, I'll try to point out a
possible shortcut.
We might observe and use the fact that
$$a\sqrt{a}+b\sqrt{b}\ge a\sqrt{b}+b\sqrt{a}$$
because
$$(a-b)(\sqrt{a}-\sqrt{b})\ge0$$
$$ \sum_{m=1}^{99} \frac{1}{(m+1)^{3/2} + m^{3/2}} < \sum_{m=1}^{99} \left(\frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right) = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{99+1}} = \frac{9}{10} $$ The above inequality is true since: $$ \begin{eqnarray} \frac{1}{(m+1)^{3/2} + m^{3/2}} &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \left( \frac{m+1}{\sqrt{m}} + \frac{m}{\sqrt{m+1}} \right)^{-1} \\ &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \left( \sqrt{m} + \sqrt{m+1} + \frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right)^{-1} \\ &<& \frac{1}{\sqrt{m}\sqrt{m+1}} \frac{1}{ \sqrt{m} + \sqrt{m+1} } \\ &=& \frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \end{eqnarray} $$
