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
On
As $\sqrt n>0,$
$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}\iff2\sqrt{n(n+1)}-2n<1$$ $$\iff2\sqrt{n(n+1)}<2n+1$$
Squaring we get $$4(n^2+n)<(2n+1)^2=4n^2+4n+1\iff 1>0$$ which is true.
Can you follow the same method for the other inequality?
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$$2\sqrt{n+1}-2\sqrt{n}=2\left(\sqrt{n+1}-\sqrt{n}\right)=$$ $$=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}} \Leftrightarrow$$ $$\Leftrightarrow 2\sqrt n<\sqrt{n+1}+\sqrt{n}\Leftrightarrow\sqrt n< \sqrt{n+1}$$
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\begin{align*} 2\sqrt{n+1}-2\sqrt{n} &= 2\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})} \\ &= 2\frac{1}{(\sqrt{n+1}+\sqrt{n})} \\ &< \frac{2}{2\sqrt{n}} \text{ since } \sqrt{n+1} > \sqrt{n}\\ &=\frac{1}{\sqrt{n}} \end{align*} Similar proof for the other inequality.
On
$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}\iff\sqrt{n^2+n}-n\lt\frac 12\lt n-\sqrt{n^2-1}$$ Put $X=n+\frac 12$ and $Y=n-\frac 12$ so you get the evidences $$\sqrt{X^2-\frac 14}\lt X\iff X^2-\frac 14\lt X^2$$ and $$\sqrt{Y^2-\frac 14}\lt Y\iff Y^2-\frac 14\lt Y^2$$
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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.
$f(x)=\frac{1}{\sqrt{x}}$ is a decreasing function on $\mathbb{R}^+$, hence: $$ 2\sqrt{n+1}-2\sqrt{n}= \int_{n}^{n+1}\frac{dx}{\sqrt{x}} \leq \frac{1}{\sqrt{n}} $$ as well as $$ 2\sqrt{n}-2\sqrt{n-1}=\int_{n-1}^{n}\frac{dx}{\sqrt{x}}\geq \frac{1}{\sqrt{n}}.$$