Prove this problem using Mean - Value theorem let $m$ and $n$ are in $\mathbb{N}$ with $m>n$ . Use the Mean Value Theorem, to prove that  $(1+x)^\frac{m}{n}\ge 1+\frac{m}{n}x, x \ge -1$
since $(1+x)^\frac{m}{n}$  is continuous on $[-1,x]$ and differentiable on $(-1,x)$ so by mean value theorem $\frac{f(x)-f(-1)}{x+1}=f'(c)$ for $-1\le c\le x$
then $\frac{(1+x)^{\frac{m}{n}}}{x+1}=\frac{m}{n}(1+c)^{\frac{m}{n}-1}$
form here how to we processed

Let $f(x) = (1+x)^{\frac mn}$. Then
$$f'(x) = \frac mn (1+x)^{\frac mn -1} =\frac mn (1+x)^{\frac {m-n}n} $$
$f'(x)$ is increasing for $x\ge -1$, and in particular, $f'(0) = \frac mn$.
Case 1: $x > 0$. Consider $f(0)$ and $f(x)$. By mean value theorem, there exists a $c\in (0, x)$ that satisfies
$$\begin{align*}
f'(c) &= \frac{f(x)-f(0)}{x-0}\\
&= \frac{(1+x)^{\frac mn}-1}{x}\\
\end{align*}$$
Since $f'(x)$ is increasing for $x > -1$,
$$\begin{align*}
f'(c) &\ge f'(0)\\
\frac{(1+x)^{\frac mn}-1}{x} &\ge \frac mn\\
(1+x)^{\frac mn} &\ge 1 + \frac mn x
\end{align*}$$

Case 2: $-1 \le x < 0$. Consider $f(x)$ and $f(0)$, and there exists a $c\in(x, 0)$ that satisfies
$$\begin{align*}
f'(c) &= \frac{f(0)-f(x)}{0-x}\\
&= \frac{1-(1+x)^{\frac mn}}{-x}\\
\end{align*}$$
Since $f'(x)$ is increasing for $x > -1$,
$$\begin{align*}
f'(c) &\le f'(0)\\
\frac{1-(1+x)^{\frac mn}}{-x} &\le \frac mn\\
1-(1+x)^{\frac mn} &\le -\frac mn x\\
(1+x)^{\frac mn} &\ge 1 + \frac mn x
\end{align*}$$

Case 3: $x=0$ is trivial.
