An inequality for all natural numbers Prove, using the principle of induction, that for all $n \in \mathbb{N}$, we have have the following inequality:
$$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$$

Try this:
The inequality hold for $k=1$, assume it holds for $k=n$. Now look at $k=n+1$:
$$
S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}
$$ 
The last step is the assumption. So LHS is $2\sqrt{n} + \frac{1}{\sqrt{n+1}}$ and RHS is $2 \sqrt{n+1}$ (induction). Clearly 
$$
2 \sqrt{n+1} - 2 \sqrt{n}=2(\sqrt{n+1} -  \sqrt{n})=\frac{2}{\sqrt{n+1}+\sqrt{n}}
>\frac{1}{\sqrt{n+1}}
$$
Therefore, 
$$
S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}<2 \sqrt{n+1}
$$
