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

Note that $\frac{(1+x)^\frac{m}{n}-1}{x}=\frac{m}{n} {(1+c)}^{\frac{m}{n}-1}$ for some $c$ in between $x$ and $0$.
For $x>0$, ${(1+c)}^{\frac{m}{n}-1}>1$; for $x<0$, ${(1+c)}^{\frac{m}{n}-1}<1$.
