<s>[INST] 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!
 [/INST] For the base case, you should prove that
$$
1+\frac{1}{\sqrt{2}}>\sqrt{2}
$$
which is equivalent to $\sqrt{2}+1>2$ or $\sqrt{2}>1$; since $2>1$, we are done.
Now assume that the statement is true for $n$; you need to prove that
$$
\sum_{i=1}^{n+1}\frac{1}{\sqrt{i}}>\sqrt{n+1}
$$
By the induction hypothesis, you know that the left-hand side is greater than
$$
\sqrt{n}+\frac{1}{\sqrt{n+1}}
$$
so it's sufficient to show that
$$
\sqrt{n}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1}
$$
which is equivalent to
$$
\sqrt{n(n+1)}+1>n+1
$$
Can you finish?
</s>