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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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