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:
$$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$$
 A: HINT
First prove that $2 \sqrt{n} + \dfrac1{\sqrt{n+1}} < 2\sqrt{n+1}$.
To prove this note that $$\sqrt{n+1} - \sqrt{n} = \dfrac1{\sqrt{n+1} + \sqrt{n}} > \dfrac1{2\sqrt{n+1}}$$
Now couple this with what you have at your induction step.
A: 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).
A: 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, 
$$
S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}<2 \sqrt{n+1}
$$
