Prove the inequality - inequality involving surds Prove that for $r$ greater than or equal to 1:
$\displaystyle 2(\sqrt{r+1} - \sqrt{r}) < \frac{1}{\sqrt{r}} < 2(\sqrt{r}-\sqrt{r-1})$
Any help on this would be much appreciated.

Let $f(x) = \sqrt x$. By mean value theorem, there exist $q\in (r-1,r)$ and $s \in (r, r+1)$ that satisfy
$$f(r) - f(r-1) = f'(q),\quad f(r+1)-f(r) = f'(s).$$
By the concavity of $f$, since $q<r<s$,
$$\begin{array}{rcl}
f'(q) >& f'(r) &> f'(s)\\
f(r)-f(r-1) >& f'(r) &> f(r+1) - f(r)\\
\sqrt{r}-\sqrt{r-1} >& \dfrac1{2\sqrt r} &> \sqrt{r+1} - \sqrt{r}\\
2(\sqrt{r+1}-\sqrt{r}) <& \dfrac1{\sqrt r} &< 2(\sqrt{r} - \sqrt{r-1})
\end{array}$$
