Prove $\frac{m(1-m)}{n}+1\leq\frac{n^m-m^m}{x^m}\leq m(1-m)+n$

Question

Let $f(x)=\frac{1}{m}x^m+\frac{1}{n}-x$,such that $\frac{1}{m}+\frac{1}{n}=1(n>m>1),x\in[m,n],f(x)\geq0 $

Prove that:

$$\frac{m(1-m)}{n}+1\leq\frac{n^m-m^m}{x^m}\leq m(1-m)+n$$

This is a question of Unified National Graduate Entrance Examination(China). I don't know how to do this problem. So I want to get some help and hint. If I make progress，I will edit it soon.

Thanks. Any help would be greatly appreciated.

Answer 1

Hint: We have $$\frac{f(n)-f(m)}{n-m}=\frac{\xi^{m-1}m}{m}-1$$ so $$\frac{\frac{n^m-m^m}{m}+m-n}{(n-m)}=\frac{n^m-m^m}{m(n-m)}-1$$ and use that $$n=\frac{m}{m-1}$$ Doing this we get $$n^m-m^m\le x^{m-1}m(n-m)\le x^mm(n-m)$$ (since $$x>1$$) So we get $$\frac{n^m-m^m}{x^m}\le m(n-m)=m(1-m)+n$$ we have only to show that $$mn\le m+n$$ this is (substituting $$n=\frac{m}{m-1}$$) we get $$\frac{m}{m-1}\le 1 +\frac{1}{m-1}$$ this is true since $$m\le m-1+1$$ Similary we can show the lower bound! ($\xi \in [m,n]$)

We have $$\frac{n^m-m^m}{m(n-m)}=\xi^{m-1}\le x^{m-1}\le x^m$$ since $$\xi \in [m,n]$$ and $$x>1$$