Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ 
Possible Duplicate:
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction 

How do I prove the following?
$$\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}},$$ for all $n \in\mathbb{Z}$, $n\ge 2$.
 A: For any $r \in [1, n]$, we have
$$
\frac{1}{\sqrt{r}} \geqslant \frac{1}{\sqrt{n}}.
$$
with strict inequality for $1 \leqslant r \lt n$. Adding all these $n$ inequalities,
$$
\sum_{r=1}^{n} \frac{1}{\sqrt{r}} \gt n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}.
$$

Proof using induction. The OP requested a proof using induction. I am assuming you can handle the base case $n=2$. 
Now, assume that the inequality holds for some $n \geqslant 2$; we will verify the inequality for $n+1$: 
$$
\begin{align*}
\sum_{r=1}^{n+1} \frac{1}{\sqrt{r}} 
&= \sum_{r=1}^{n} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{n+1}}  
\\ &\gt \sqrt{n} + \frac{1}{\sqrt{n+1}} 
\\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - (\sqrt{n+1} - \sqrt{n}) 
\\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+1} + \sqrt{n}} 
\\ &\gt \sqrt{n+1} ,
\end{align*}
$$
which is what we want to show. 
Notice that out second inequality is a bit too crude. robjohn's answer shows how to get a better bound by being more careful. 
A: We can even do better:
$$
\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}<\frac{1}{2\sqrt{n-1}}\tag{1}
$$
Summing, we get
$$
\sqrt{n}-1<\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{2}
$$
So, for $n\ge2$,
$$
\sqrt{n}<1+\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{3}
$$
which is a better bound for every $n$.
