Q:

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

A:

Cauchy's Mean Value Theorem says that for some $\xi$ between $0$ and $x$,
$$
\frac{(1+x)^p-1}{(1+px)-1}=\frac{p(1+\xi)^{p-1}}{p}=(1+\xi)^{p-1}
$$
Now if $x\ge0$, we have
$$
\frac{(1+x)^p-1}{(1+px)-1}=\frac{p(1+\xi)^{p-1}}{p}\ge1
$$
which leads to
$$
(1+x)^p\ge1+px
$$
On the other hand, if $-1\le x\le0$, then
$$
\frac{(1+x)^p-1}{(1+px)-1}=\frac{p(1+\xi)^{p-1}}{p}\le1
$$
but since $(1+px)-1\le0$, we have again
$$
(1+x)^p\ge1+px
$$

