<s>[INST] Prove $ \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2} $ by induction 
Prove by induction that for all $n > 0$,
$$
\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots
+ \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}
$$

I have done the basis step, where $n = 1$ and showed that L.H.S is > R.H.S
For inductive step, I assume $n = k$ & L.H.S > R.H.S is true. 
However, I am stuck at showing how for $k + 1$, it is also true for L.H.S > R.H.S
Any help will be much appreciated. Thanks!
 [/INST] By cross-multiplying, it can easily be seen that
$$
\begin{align}
\frac{\sqrt{k+1}}{2k}
&\gt\frac1{\sqrt{k+1}+\sqrt{k\,}}\\[6pt]
&=\sqrt{k+1}-\sqrt{k\,}
\end{align}
$$
Summing gives
$$
\sum_{k=1}^n\frac{\sqrt{k+1}}{2k}
\gt\sqrt{n+1}-1
$$
This is not the inequality given, but for $n\ge2$, $\sqrt{n+1}-1\gt\frac{\sqrt{n}}2$.
</s>