# Prove that $2\sqrt{n}\sqrt{n+1} < 2n + 1$ for all positive integers.

I've been testing this with many values and it seems to always be true. I've been trying to rework the inequality into a form where it's much more obvious that the left hand side is always less than the right, but can't seem to do it. Can anyone help me out here?

Thanks.

• Generalise to arbitrary numbers: $2ab \leqslant a^2 + b^2$. – Daniel Fischer Oct 9 '14 at 18:57
• square both sides – John Oct 9 '14 at 18:58

## 3 Answers

$$2\sqrt{n}\sqrt{n+1}=\sqrt{4n^2+4n}<\sqrt{4n^2+4n+1}=2n+1.$$ Note. This inequality holds for every non-negative real $n$ (not only integer.)

• Thanks, that's exactly what I was looking for! – user176049 Oct 9 '14 at 19:00
• It's worth noting that this proves it for all $n\ge0$ not just integers – Alice Ryhl Oct 9 '14 at 19:01
• That is in fact right! – Yiorgos S. Smyrlis Oct 9 '14 at 19:01

By the AM-GM inequality, $\displaystyle\sqrt{n}\sqrt{n+1}\le\frac{2n+1}{2}$.

• This should be a standard tool in any inequality toolbox. – Steven Stadnicki Oct 13 '14 at 21:58

$$2n+1 - 2\sqrt n\sqrt{n+1}=(\underbrace{\sqrt n-\sqrt{n+1}}_{\ne 0})^2 >0$$