How to get $\sqrt {k} + \frac{1}{\sqrt{k+1}}$ in the form $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$?

Question

I was wondering if it is possible to get $\sqrt {k} + \dfrac{1}{\sqrt{k+1}}$ in the form $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$, and if so, how?

I ask this, because I'm following this answer, and I get lost at how they arrive at $\dfrac{\sqrt{k^2} + 1}{\sqrt{k+1}}$.

Thanks.

EDIT: To further clarify what I'm not understanding, please read below:

I don't understand how the author goes from the line:

$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}$

to the line:

$\frac{\sqrt{k^2 + k} + 1}{\sqrt{k + 1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}}$

I would really appreciate it if someone could lead me through to that line and explain exactly what is being done on each line.

Thanks again.

Answer 1

That answer says

$$\sqrt {k} + \frac{1}{\sqrt{k+1}}= \frac{(\sqrt {k})(\sqrt {k+1}) +1}{\sqrt{k+1}}=\frac{\sqrt {k(k+1}) +1}{\sqrt{k+1}}=\frac{\sqrt {k^2+\color{red}{k}} +1}{\sqrt{k+1}}\color{blue}{\ge}\frac{\sqrt {k^2} +1}{\sqrt{k+1}}$$

It's just like saying $\sqrt{a+b}\ge\sqrt{a}$ for $b\ge0$

Answer to Edit:

Author want to show that $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}$$ So you first assume That it is true for $n=k$ that is $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} +...+ \frac{1}{\sqrt{k}} > \sqrt{k}$$ And then You add $\dfrac{1}{\sqrt{k+1}}$ to both sides and then show that it is also true for $n=k+1$

Answer 2

It cannot hold obviously. When $k=\color{red}1$, $\sqrt {k} + \frac{1}{\sqrt{k+1}}=\color{red}{1+\frac{\sqrt2}{2}}$, however, $\frac{\sqrt{k^2} + 1}{\sqrt{k+1}}=\color{red}{\sqrt2}.$