EDIT: Why is $\sum_{k=1}^n \frac{1}{\sqrt{k}}>\sqrt{n}$?

Question

Having problems with another inductive proof.. again with the algebra

$$\sum_{k=1}^n \frac{1}{\sqrt{k}}>\sqrt{n} :\ \forall n\in\mathbb{N}, n>1$$

What I've done:

1. Proved for n=1
2. Showed I need to prove $$\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}}>\sqrt{n+1}$$
3. Attempt: $$\sum_{k=1}^{n} \frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}>\sqrt{n}+\frac{1}{\sqrt{n+1}}$$ $$\iff \sum_{k=1}^{n+1} \frac{1}{\sqrt{k}}>\sqrt{n}+\frac{1}{\sqrt{n+1}} $$

Again, I come no futher than this. I assume its something I don't know or can't see.

Cheers!

Andy

Edit: After being referred to another similar question, what I don't understand is why $$\sqrt{n}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1}$$ which it obviously must be.

In the same vein, during my proof I didn't know (algebraically) to prove $$\sum_{k=1}^n \frac{1}{\sqrt{k}}>\sqrt{n}$$

If I put numbers in I can see that it is trivially true but not why it is generally true..