discrete mathematics - 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!

Answer

(Someone please check for error ... I'm not confident if I got this correct)

Since for the base case $n=1$: $\dfrac{\sqrt2}2>\dfrac{\sqrt 1}2$, you just have to prove that $\dfrac{\sqrt{n+1}}{2n}\geq\dfrac{\sqrt{n}}2-\dfrac{\sqrt{n-1}}2$ for $n>1$ since, if $a>b$ then $a+c>b+d$ for $c\geq d$.

$$\begin{align*} \frac{\sqrt{n+1}}{2n}&\geq\frac{\sqrt n-\sqrt{n-1}}2\\ \\ \frac{\sqrt{n+1}}{n}&\geq\sqrt n-\sqrt{n-1}\\ \\ \frac{\sqrt{n+1}}{\sqrt n-\sqrt{n-1}}&\geq n\\ \\ \sqrt{n+1}(\sqrt n+\sqrt{n-1})&\geq n\\ \\ \sqrt{n^2+n}+\sqrt{n^2-1}&\geq n\\\square \end{align*}$$

(note that $\sqrt{n^2+n}>n$ since $\sqrt{n^2}=n$ for $n\geq1$)