Proving $\sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n}$ with induction Problem: Prove with induction that \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n} \end{align*} for every natural number $n \geq 2$.
Attempt at proof: Basic step:  For $n = 2$ we have $1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ which is correct.
Induction step: Suppose the assertion holds for some natural number $n$, with $n > 2$. Then we now want to prove that it also holds for $n +1$, i.e. that \begin{align*} \sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} > \sqrt{n+1} \end{align*}
Now we have that \begin{align*} \sum_{j=1}^{n+1} \frac{1}{\sqrt{j}}  = \sum_{j=1}^n \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{n+1}} > \sqrt{n} + \frac{1}{\sqrt{n+1}} \end{align*} or \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{n+1}} > \frac{\sqrt{n} \sqrt{n+1} + 1}{\sqrt{n+1}} \end{align*}
Now I'm stuck, and I don't know how to get $\sqrt{n+1}$ on the right hand side. Help would be appreciated.

Notice that : 
$ \sqrt{n}\sqrt{n+1} + 1 > \sqrt{n}\sqrt{n} + 1 $
$\sqrt{n}\sqrt{n+1} + 1 > n + 1 $
$ \sqrt{n}\sqrt{n+1} + 1 > \sqrt{n+1}\sqrt{n+1} $
$ \frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}} > \sqrt{n+1}. $
