Inequality proof using real numbers with describe

$$ \sqrt{k+1}-\sqrt{k} = \frac{1}{\sqrt{k+1}+\sqrt{k}} < \frac{1}{2\sqrt{k}} $$ hence by summing these inequalities for $k=1,2,\ldots,n$ we get: $$ \sqrt{n+1}-\sqrt{1} < \frac{1}{2}\left(\frac{1}{\sqrt{1}}+\ldots+\frac{1}{\sqrt{n}}\right)$$ as wanted.

You can do this by induction.

The base case is easy.

By induction it suffices to show that $2(\sqrt{n+1}-1)+\frac{1}{\sqrt{n+1}} \ge 2(\sqrt{n+2}-1)$. Rearranging this, we see that this is equivalent to $\frac{1}{\sqrt{n+1}} \ge 2(\sqrt{n+2}-\sqrt{n+1})$. We can rewrite the right-hand side of this using the formula $a^2-b^2=(a+b)(a-b)$: so the above is equivalent to $\frac{1}{\sqrt{n+1}} \ge \frac{2[(n+2)-(n+1)]}{\sqrt{n+2}+\sqrt{n+1}}$, i.e. $\frac{1}{\sqrt{n+1}} \ge \frac{2}{\sqrt{n+2}+\sqrt{n+1}}$. It should be easy now to see why this inequality holds, and so we're done.