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.
 A: 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.
A: 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*}
A: $n^2+n=(n+\frac12)^2-\frac14<(n+\frac12)^2\Rightarrow \sqrt{n^2+n}-n<\frac12$
$\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}>\frac{n}{n+\frac12+n}>\frac{n}{2n+1}$
A: More an observation than an answer since the OP explicitly asks for applying the Mean value Theorem.
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.
