# How to prove the inequality $\sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}$ for $n\in\mathbb{Z}^+$?

I have to prove this inequality: $$\forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}$$

So far, I have done the base cases and assumed the inequality is true for some integer k, and I have gotten to the point where I need to show that: $$\frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} < \frac{\sqrt{k+1}}{2}$$ Can somebody point me in the right direction? I have tried squaring both sides, but because there is a + sign on the LHS, there are still square roots and it only makes things more complicated.

Is there a name for this inequality?

• Thanks for the help, but I am finding it hard to understand what you guys are talking about. If it is possible, please go step by step, forgive me if I am asking too much. Is there a name for this inequality? Oct 24 '13 at 3:23

I am not a fan of using (in)equality that you have to prove, since you have to be really careful what you do with that. So, let's try this:

\begin{align*} \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} &> \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+4} = \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2(k+2)} \\ &> \frac{\sqrt{k}}{2} + \frac{1}{2\sqrt{k+2}} = \frac{1}{2} \left(\sqrt{k} + \frac{1}{\sqrt{k+2}} \right) = \frac{1}{2} \sqrt{ k + 2\sqrt{\frac{k}{k+2}} + \frac{1}{k+2} } \\ &> \frac{1}{2} \sqrt{k + 1} \end{align*}

You'll need to justify why

$$2\sqrt{\frac{k}{k+2}} + \frac{1}{k+2} > 1,$$

But that should be really easy.

I would look at your inequality this way: $$\frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}-\frac{\sqrt{k}}{2} .$$ The right hand side is $\frac1{2(\sqrt{k+1}+\sqrt{k})}$. So we have $$\frac1{2(\sqrt{k+1}+\sqrt{k})}<\frac1{4\sqrt{k}}<\frac1{2\sqrt{k+1}}=\frac{\sqrt{k+1}}{2(k+1)}<\frac{\sqrt{k+2}}{2(k+1)}.$$

• I get your $$\frac{\sqrt{k+2}}{2k+2} > \frac{\sqrt{k+1}}{2}-\frac{\sqrt{k}}{2} .$$ part but I do not understand the part where you change it to $$\frac1{2(\sqrt{k+1}+\sqrt{k})}$$ Oct 24 '13 at 1:19
• @mib1413456 After combining the right into one fraction over $2$, multiply top and bottom by the conjugate $\sqrt{k+1}+\sqrt{k}.$ Oct 24 '13 at 1:46
• May I know how did you arrive at $$\frac1{2(\sqrt{k+1}+\sqrt{k})}<\frac1{4\sqrt{k}}<\frac1{2\sqrt{k+1}}=\frac{\sqrt{k+1}}{2(k+1)}<\frac{\sqrt{k+2}}{2(k+1)}.$$ So far, I only got $$\frac{\sqrt{k+2}}{2k+2} > \frac1{2(\sqrt{k+1}+\sqrt{k})}$$ Oct 24 '13 at 2:44
• This last inequality is the one you want to prove. What I did (to prove that inequality) is, in order of the inequality signs: 1) first $\sqrt{k+1}>\sqrt{k}$; 2) $2\sqrt{k}\geq\sqrt{k+1}$ (this is equivalent to $\sqrt{1+1/k}\leq2$); 3) $1/\sqrt{k+1}=(k+1)/\sqrt{k+1}$; 4) $\sqrt{k+1}<\sqrt{k+2}$. Oct 24 '13 at 2:52


\begin{align} {\large\sum_{i = 1}^{n}{\root{\vphantom{\large A}i + 1} \over 2i}} &= {1 \over 2}\sum_{i = 1}^{n}\pars{{1 \over \root{\vphantom{\large A}i + 1}} + {1 \over i\root{\vphantom{\large A}i + 1}}} > {1 \over 2}\sum_{i = 1}^{n}\pars{{1 \over \root{\vphantom{\large A}i + 1}} + {1 \over n\root{\vphantom{\large A}i + 1}}} \\[3mm]&= {1 \over 2}\,{n + 1 \over n}\sum_{i = 1}^{n} {1 \over \root{\vphantom{\large A}i + 1}} > {1 \over 2}\,{n + 1 \over n}\pars{n\,{1 \over \root{\vphantom{\large A}n + 1}}} \\[3mm]&={\large% {\root{n + 1} \over 2} > {\root{n} \over 2}} \end{align}