Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.
$$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$
Perhaps someone can help me out I don't understand how to move forward from here:
$$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$
proof and explanation would greatly be appreciated :)
Thanks :)
EDIT sorry meant GE not = fixed :)

For those who strive for non-induction proofs...
Since $\frac 1{\sqrt k} \ge \frac 1{\sqrt n}$ for $1 \le k \le n$, we actually have
$$
\sum_{i=1}^n \frac 1{\sqrt k} \ge \sum_{i=1}^n \frac 1{\sqrt n} = \frac n{\sqrt n} = \sqrt n.
$$
