Question
If n\geq 1 Proof of the square root inequality
2\sqrt{n+1}-2\sqrt{n}<\dfrac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}
Answer
Since f(x) = \sqrt{x} is an increasing function,
\sqrt{n+1} > \sqrt{n}>\sqrt{n-1}
Add \sqrt{n} in each side, which will not change the direction of the inequality sign
\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}>\sqrt{n-1}+ \sqrt{n}
(\sqrt{n+1} + \sqrt{n})\cdotp \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} > 2\sqrt{n}>(\sqrt{n-1}+ \sqrt{n})\cdotp \dfrac{\sqrt{n-1}- \sqrt{n}}{\sqrt{n-1}- \sqrt{n}}
\dfrac{1}{\sqrt{n+1} - \sqrt{n}} > 2\sqrt{n}>\dfrac{1}{\sqrt{n}-\sqrt{n-1} }
Taking the reciprocal in the inequality leads to change of the inequality sign.
\sqrt{n+1} - \sqrt{n} <\dfrac{1}{2\sqrt{n} }<\sqrt{n}-\sqrt{n-1}
Therefore,
2\sqrt{n+1} -2 \sqrt{n} <\dfrac{1}{\sqrt{n} }<2\sqrt{n}-2\sqrt{n-1}