Showing that $a\le\frac{1}{2\sqrt{n}}$

Question

Let: $$a= \sqrt{n+1}-\sqrt{n}$$ such that n is a natural number. The problem is to show that: $$a\le\frac{1}{2\sqrt{n}}$$. My approach: I could show that: $\frac{1}{2\sqrt{n}}\ge\frac{\sqrt{n+1}}{2}$, so if I could show that $$a\le\frac{\sqrt{n+1}}{2}$$ it'd be done. Any hint or help are much appreciated

Answer 1

You have: $$ a=\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}$$ Since $\sqrt{n+1}+\sqrt n\ge2\sqrt n$, you get $a\le \frac1{2\sqrt n}$.

Answer 2

By the mean value theorem, there exists $c \in (n, n+1)$ such that $\sqrt{n+1} - \sqrt{n} = \frac{1}{2 \sqrt{c}} \leq \frac{1}{2 \sqrt{n}}$ where the last inequality comes from the fact that $\sqrt{.}$ is non-decreasing.