# Proving $\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$

I would like to prove using Mean Value theorem for $$n \ge 1$$ $$\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$$

RHS can be proved by rationalizing the square root term, not sure about the LHS.

Define $$f(x)=\sqrt x$$ and apply MVT on the interval $$[n^2,n^2+n]$$. You get $$\sqrt{n^2+n}-n=\frac{n}{2\sqrt{n^2+\xi n}}$$ where $$0<\xi<1$$. Now play with inequalities.

$$n^2+n=(n+\frac12)^2-\frac14<(n+\frac12)^2\Rightarrow \sqrt{n^2+n}-n<\frac12$$

$$\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}>\frac{n}{n+\frac12+n}>\frac{n}{2n+1}$$

$$\begin{eqnarray*} \sqrt{n^2+n} -n & = & \frac{n^2 + n - n^2 }{\sqrt{n^2+n} + n} \\ & \color{blue}{>} & \frac{n}{\sqrt{n^2+2n+1}+n} \\ & = & \frac{n}{\sqrt{(n+1)^2}+n} \\ & = & \frac{n}{2n+1} \\ \end{eqnarray*}$$
Write the given inequality chain as $$\frac2{\frac1n+\frac1{n+1}}-n \:<\: \sqrt{n(n+1)}-n \:<\: \frac{n+(n+1)}2-n\,,$$ then it directly follows from the Harmonic-Geometric-Arithmetic mean inequality.
Instead of "$$\leqslant$$" the stronger "$$<$$" holds as the variables $$\,n\,$$ and $$\,n+1\,$$ are not equal.