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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.

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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.

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

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

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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*}

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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.

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  • $\begingroup$ That's really beautiful. $\endgroup$ – Goal123 Feb 6 at 8:25

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