Prove the inequality - inequality involving surds

Question

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.

Answer 1

Hint: $\frac{1}{\sqrt{r}}=\frac{2}{\sqrt{r}+\sqrt{r}}$ and $r-1<r<r+1$.

Answer 2

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 $q<r<s$, $$\begin{array}{rcl} f'(q) >& f'(r) &> f'(s)\\ f(r)-f(r-1) >& f'(r) &> f(r+1) - f(r)\\ \sqrt{r}-\sqrt{r-1} >& \dfrac1{2\sqrt r} &> \sqrt{r+1} - \sqrt{r}\\ 2(\sqrt{r+1}-\sqrt{r}) <& \dfrac1{\sqrt r} &< 2(\sqrt{r} - \sqrt{r-1}) \end{array}$$

Answer 3

Let $g(x) = \frac1{\sqrt x}$, and $g(x)$ is strictly decreasing. Then this inequality holds:

$$\int_{r-1}^rg(x)\ dx > 1\cdot g(r) > \int_{r}^{r+1}g(x)\ dx\\ \int_{r-1}^r\frac{dx}{\sqrt x} > \frac{1}{\sqrt r} > \int_{r}^{r+1}\frac{dx}{\sqrt x}\\ 2(\sqrt r - \sqrt{r-1}) > \frac{1}{\sqrt r} > 2(\sqrt{r+1} - \sqrt r)$$