Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction

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,

A very short (though non-inductive) proof:

$$ \sum_{k=1}^n \frac{1}{\sqrt{k}} \ge \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sum_{k=1}^n \frac{\sqrt{k} - \sqrt{k-1}}{(\sqrt{k} + \sqrt{k-1})(\sqrt{k} - \sqrt{k-1})} = \sum_{k=1}^n (\sqrt{k} - \sqrt{k-1}) = \sqrt{n} $$

I won't use induction:

On the left side you have a sum with $n$ terms, the smallest one is $\frac{1}{\sqrt{n}}$. So you get the inequality:

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}\ge \frac{1}{\sqrt{1}}+(n-1)\frac{1}{\sqrt{n}}=\left(1-\frac{1}{\sqrt{n}}\right)+\sqrt{n}$$

And now you can see easily that the right hand side is larger than $\sqrt{n}$, for all $n>1$.

I hope this helps.