Prove inequality by induction: $\sum _{i=1}^n\frac{1}{\sqrt{i}} > 2\left(\sqrt{n+1}-1\right)$ for $n \ge 1$

Question

I worked through most of this problem but am getting stuck on the final bit.

Base: $n = 1$: $\frac{1}{\sqrt{1}} > 2(\sqrt{2}-1)\to\sqrt{9}> \sqrt{8}$. through moving stuff around

Hypothesis: true for $n = k$, $\sum _{i=1}^k\frac{1}{\sqrt{i}}>2\sqrt{k+1}-1$

Induction: for $n = k+1$, $\sum _{i=1}^k\frac{1}{\sqrt{i}}+\frac{1}{\sqrt{k+1}}>2\left(\sqrt{\left(k+1\right)+1}-1\right) = 2\left(\sqrt{k+1}-1\right)+\frac{1}{\sqrt{k+1}}$

I've tried all kinds of things after this, but can't seem to get it to the right spot. Anyone have insight?

Answer 1

Where $S(k+1)$ is the sum of $k+1$ terms, you have

$S(k+1)>2(\sqrt{k+1}-1)+\dfrac{1}{\sqrt{k+1}}.$

Now render

$\dfrac{1}{\sqrt{k+1}}=\dfrac{2}{\sqrt{k+1}+\sqrt{k+1}}>\color{blue}{\dfrac{2}{\sqrt{k+2}+\sqrt{k+1}}}$

and multiply the numerator and denominator of the blue expression by the appropriate conjugate. What happens then?