# Prove $\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$

What would you suggest for the following inequality? $$\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$$

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 $$(a-b)(\sqrt{a}-\sqrt{b})\ge0$$

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According to Mathematica the actual sum is approximately 0.860068, so the inequality is pretty sharp. – Grumpy Parsnip Feb 27 '13 at 19:55
@JimConant try to remove N[] from Sum[] and you'll see that Mathematica actually introduced a lot of extra terms :) – Kaster Feb 27 '13 at 20:02
@Kaster: are you claiming the Mathematica calculation is wrong? – Grumpy Parsnip Feb 27 '13 at 20:18
@JimConant I thought so, but I think it's just some simplifications, like $\frac 1{343 + 250 \sqrt 2} = \frac 1{50^{\frac 32} + 49^{\frac 32}}$. My bad. – Kaster Feb 27 '13 at 20:25
Chris, please avoid using titles like "Interesting $X$". They are non informative and subjective. – Pedro Tamaroff Feb 28 '13 at 18:40

$$\sum_{m=1}^{99} \frac{1}{(m+1)^{3/2} + m^{3/2}} < \sum_{m=1}^{99} \left(\frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right) = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{99+1}} = \frac{9}{10}$$ The above inequality is true since: $$\begin{eqnarray} \frac{1}{(m+1)^{3/2} + m^{3/2}} &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \left( \frac{m+1}{\sqrt{m}} + \frac{m}{\sqrt{m+1}} \right)^{-1} \\ &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \left( \sqrt{m} + \sqrt{m+1} + \frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right)^{-1} \\ &<& \frac{1}{\sqrt{m}\sqrt{m+1}} \frac{1}{ \sqrt{m} + \sqrt{m+1} } \\ &=& \frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \end{eqnarray}$$