doubt with unequal induction Show that for all n ≥ 2 it is verified:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+ ...+\frac{1}{\sqrt{n}} > \sqrt{n} $$
I have used the induction method to arrive at the following result:
$$\sqrt{k}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1} $$
someone can help me, I think I have not solved it well
 A: That ... is enough.
If you have proven that $\sqrt{k} + \frac {1}{\sqrt{k+1}} > \sqrt{k+1}$
Then you can state that for $\frac 1{\sqrt 1} + \frac 1{\sqrt 2} =\sqrt{1} + \frac {1}{\sqrt{2}} > \sqrt{2}$, so the base step of the induction is done.
You can state if you know $\frac{1}{\sqrt{1}} + .... + \frac{1}{\sqrt{j}} > \sqrt j$
Then $(\frac{1}{\sqrt{1}} + .... + \frac{1}{\sqrt{j}}) + \frac 1{\sqrt{j+1}} >$
$(\sqrt{j})+ \frac 1{\sqrt{j+1}}>$
$\sqrt {j+1}$.
So the induction step holds.
So it has been successfully proven by induction.
A: Although the following is not strictly based on induction (i.e., other than use of telescoping series), I thought it might be instructive to present a direct proof of the coveted inequality.
Note that we have
$$\begin{align}
\sqrt{n}&=\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)\tag{telescoping  series}\\\\
&=\sum_{k=1}^n\frac{1}{\sqrt{k}+\sqrt{k-1}}\tag{rationalizing numerator}\\\\
&<\sum_{k=1}^n \frac1{\sqrt k}\tag{For $k>1$, $\sqrt{k-1}> 0$}
\end{align}$$
as was to be shown!
A: \begin{eqnarray*}
k>0\\
k^2+k> k^2 \\
\sqrt{k(k+1)} > k \\
\sqrt{k(k+1)}+1 > k+1 \\
\sqrt{k} +\frac{1}{\sqrt{k+1}}> \sqrt{k+1} \\
\end{eqnarray*}
