Show: $n(x-y)y^{n-1} < x^n-y^n < n(x-y)x^{n-1}$

Question

Given $1<n\in \mathbb{N}$ and $0\leq y <x$ where $x, y\in \mathbb{F}$ ($\mathbb{F}$ is the ordered field)

It's been proven that $x^n-y^n=(x-y)\sum_{i=1}^{n} x^{n-i}y^{i-1}$

Show: $n(x-y)y^{n-1} < x^n-y^n < n(x-y)x^{n-1}$

Progress made: $ny^{n-1} < x^{n-1}+x^{n-2}y+,\dots,+xy^{n-2}+y^{n-1}<nx^{n-1}$

I can understand $y^{n-1}<nx^{n-1}$ though what about $ny^{n-1} < x^{n-1}+x^{n-2}y+,\dots,+xy^{n-2}\dots$?

Answer 1

We have $y<x$, raise this to the power of $n-i$ and multiply by $y^{i-1}$ (and similarly for the other inequality) \begin{eqnarray*} y^{n-1} < x^{n-i}y^{i-1} < x^{n-1} \end{eqnarray*} provided $1<i<n$. Now sum over $i=1,\cdots ,n$ and the result follows.