Proof By Induction: $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ Here's what I have so far: $$\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$$ where $n\ge2$.
Test for base case
$$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} > \sqrt 2$$
Simplify
$$ 1 + .707... > 1.414... $$
$$ 1.707... > 1.414... $$
This statement is true.
Thus we can assume $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ is true for $\forall n$
Now prove $n + 1$
Expand 
$$\sum_{i=1}^n \frac{1}{\sqrt i} = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} \gt \sqrt n $$
Add the next n in the series, and add it to both sides of the inequality.
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} \gt \sqrt n + \frac{1}{\sqrt{n + 1}} $$
Rewrite
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} = \sum_{i=1}^{n + 1} \frac{1}{\sqrt i}$$
LHS is complete now we have to try to make RHS = $\sqrt {n + 1}$
$$\sqrt n + \frac{1}{\sqrt{n + 1}} = \frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}}$$
Given that n > 0
$$\frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}} = \frac{\sqrt{n(n+1)} + 1} {\sqrt{n+1}}$$
Now since this is my first induction involving inequalities or summation, I was hoping you guys could show me what the next step would be, since I'm really not sure. Thanks in advance!

You can prove this without induction. In fact, when $n=1$ it's ok; and when $n>1$, for each $i=1,..,n-1$ we have $$\frac{1}{\sqrt{i}}>\frac{1}{\sqrt{n}}$$ 
Now, $\sum\limits_{i=1}^n\frac{1}{\sqrt{i}}>n\cdot\frac{1}{\sqrt{n}}=\sqrt{n}$.
