<s>[INST] 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?
 [/INST] 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.
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