Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. For $n=2$, we have $LHS=\sqrt 2\approx1.4142$, and $RHS=\frac{1}{\sqrt 1}+\frac{1}{\sqrt 2}\approx1.7071$, thus $LHS\lt RHS$ and $P(0)$ is true.
3. Assume $P(k), k\gt2$ is true, that is $\sqrt k\lt \frac{1}{\sqrt 1}+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt k}$
4. For $n=k+1$, we have $RHS=\frac{1}{\sqrt 1}+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt {k+1}}\gt \sqrt k+\frac{1}{\sqrt {k+1}}$
5. To check if $\sqrt k+\frac{1}{\sqrt {k+1}}\gt \sqrt {k+1}$, we have
$\sqrt {k(k+1)}+1\gt k+1$ or $\sqrt {k(k+1)}\gt k$ which is true. This means that $RHS\gt \sqrt {k+1}=LHS$ in step 4.
6. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.