inequality - Proving $sumlimits_{k=1}^{n}{frac{1}{sqrt{k}}gesqrt{n}}$ 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 :)

Answer

If you wanted to prove that

$$\sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n,$$
that I can do. It is clear for $n=1$ (since we have equality then), so that it suffices to verify that
$$\sum_{k=1}^{n+1} \frac 1{\sqrt k} \ge \sqrt{n+1}$$
but this is equivalent to
$$\sum_{k=1}^{n} \frac 1{\sqrt k} + \frac 1{\sqrt{n+1}} \ge \sqrt{n+1} \ $$

and again equivalent to
$$\sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} + 1 \ge n+1$$
so we only need to prove the last statement now, using induction hypothesis. Since
$$\sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n,$$
we have
$$\sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} \ge \sqrt{n+1}\sqrt{n} \ge \sqrt{n} \sqrt{n} = n.$$
Adding the $1$'s on both sides we get what we wanted.

Hope that helps,