Estimations in a ordered field? My Problem: I am stuck with a proof strategy on the following:
So i have got an ordered field $ (K,+,*,<)  $ given. I also have 
$x,y\in K$
and $0\le y < x$
I have to proof that, for every n $\in \mathbb{N}, n \ge 2$ there is:
$ny^{n-1} < \frac{x^n-y^n}{x-y} < nx^{n-1}$
What i got so far:
It can be easily proven that
$ny^{n-1} < nx^{n-1}$. It won't help us much, i guess.
We also know that the fraction in the middle is >0 because nominator and denominator will always stay positive (because $0\le y < x$). This might be of some use.
This inequality just doesn't ring a bell. I know i have to estimate the fraction somehow. Also i wonder why the task is to prove it for a specific field, not just for $\mathbb{R}$ (What difference does it make?)
Any light you can shed on this is much appreciated.

Let $K^r$ be the real closure of the ordered field $K$. For each integer $n$, the claim
$$0 \leq y < x \implies ny^{n-1} < \frac{x^n-y^n}{x-y} < nx^{n-1} $$
is a first-order statement in the language of ordered fields. Because the theory of real closed fields is complete, this means the inequality is true for the field $K^r$ if and only if it is true for the field $\mathbb{R}$.
And because $K$ is an ordered subfield of $K^r$, if it is true for $K^r$, then it must be true for $K$ as well.
Thus, if you can prove the statement for the real numbers by your favorite means (e.g. calculus?) for all $n \geq 2$, then for each particular $n \geq 2$ we can invoke the logical machinery to conclude it must be true for all ordered fields, and thus we infer it for all $n \geq 2$ for any ordered field.

If you don't want to invoke the logical machinery to prove things automatically, you can still use the basic idea: first use calculus to prove it for $\mathbb{R}$, then digging into the $\epsilon-\delta$ definition of limits and stuff, you can often directly translate the calculus proof into pure algebra.
