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?
 A: 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.
A: 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
$$
\frac1{2(\sqrt{k+1}+\sqrt{k})}<\frac1{4\sqrt{k}}<\frac1{2\sqrt{k+1}}=\frac{\sqrt{k+1}}{2(k+1)}<\frac{\sqrt{k+2}}{2(k+1)}.
$$
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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\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}
