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 :)

A very short (though non-inductive) proof:
$$
  \sum_{k=1}^n \frac{1}{\sqrt{k}} \ge \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k-1}} = 
    \sum_{k=1}^n \frac{\sqrt{k} - \sqrt{k-1}}{(\sqrt{k} + \sqrt{k-1})(\sqrt{k} - \sqrt{k-1})} = \sum_{k=1}^n (\sqrt{k} - \sqrt{k-1}) = \sqrt{n}
$$
