Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$ I stumbled on the following inequality: For all $n\geq 1,$
$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.$$
However I cannot find the proof of this anywhere.
Any ideas how to proceed?
Edit: I posted a follow-up question about generalizations of this inequality here:
Square root inequality revisited

Let $f(x)=2\sqrt{x}$. Using mean value theorem we get
$$\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c)$$
for some $c \in (n, n+1)$. Equivalently
$$2\sqrt{n+1} - 2\sqrt{n} = \frac{1}{\sqrt c}.$$
Since $c>n$, $$\frac{1}{\sqrt c} < \frac{1}{\sqrt{n}},$$
therefore
$$2\sqrt{n+1}-2\sqrt{n} < \frac 1{\sqrt n}.$$
Right inequality can be proved in a similar manner.
