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:
$$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$$

Suppose
$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$
and
$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt {n+1}} > 2\sqrt {n+1}$
(i.e., that the induction hypothesis is false).
Subtracting these,
$\frac{1}{\sqrt {n+1}} > 2\sqrt {n+1} - 2\sqrt n
= 2(\sqrt {n+1} - \sqrt n)\frac{\sqrt {n+1} +\sqrt n}{\sqrt {n+1} +\sqrt n}
= \frac{2}{\sqrt {n+1} +\sqrt n}
$
or
$\sqrt {n+1} +\sqrt n > 2 \sqrt {n+1}$, which is false.
So the induction hypothesis is true (once a base case is established).
