Well, the induction step is as follows for $n\geq1$:
$$\frac{1}{\sqrt 1} + \ldots + \frac{1}{\sqrt n} + \frac{1}{\sqrt {n+1}}
\geq \sqrt n + \frac{1}{\sqrt {n+1}}$$
by the induction hypothesis.
Now you have to prove that for the right-hand side,
$$ \sqrt n + \frac{1}{\sqrt {n+1}}\geq \sqrt{n+1}.$$
Consider the difference of these expressions. Then
$$ \sqrt n + \frac{1}{\sqrt {n+1}} - \sqrt{n+1} =
\frac{\sqrt n\sqrt{n+1} + 1 - \sqrt {n+1}\sqrt{n+1}}{\sqrt{n+1}} =
\frac{\sqrt n\sqrt{n+1}-n}{\sqrt{n+1}}.$$
The numerator is $\sqrt n\sqrt{n+1} - n=\sqrt{n(n+1)}-n = \sqrt{n^2+n} - n\geq 0$ and so the above expression is $\geq 0$ as required.