Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$

$f(x)=\frac{1}{\sqrt{x}}$ 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 $$ 2\sqrt{n}-2\sqrt{n-1}=\int_{n-1}^{n}\frac{dx}{\sqrt{x}}\geq \frac{1}{\sqrt{n}}.$$

$$2\sqrt{n+1}-2\sqrt{n}=2\left(\sqrt{n+1}-\sqrt{n}\right)=$$ $$=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}} \Leftrightarrow$$ $$\Leftrightarrow 2\sqrt n<\sqrt{n+1}+\sqrt{n}\Leftrightarrow\sqrt n< \sqrt{n+1}$$

\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.