Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!


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

share|cite|improve this question
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
up vote 19 down vote accepted

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

share|cite|improve this answer
hehe, nice (+1) – user 1618033 Feb 27 '13 at 20:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.