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.

Why MVT? For the LHS you are asking about you have
\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*}
