Seeking for an inequality involving $(a-b)^n$ I need to estimate $(a-b)^n$ ($a, b \in \mathbb{R}, n \in \mathbb{N}$) downwards, meaning I want to find an expression $x$ such that $(a-b)^n \geq x$ where $x$ must somehow involve $a^n$. I searched a bit for such an inequality but didn't find one yet. It also should be very concise, meaning I know how to estimate it with the Binomial Theorem but I don't want to have an expression involving so many terms.
 A: Since you are explicitely stating you are interested in the dependence on $a^n$, I would write
$$
(a-b)^n = a^n (1 - b/a)^n = a^n \cdot ({\rm sign}(1-b/a))^n \cdot \exp (n \log|1-b/a|)
$$
The sign-term comes in since you have $a, b \in \mathbb{R}$, which leaves all options open.
Now if $a^n \cdot ({\rm sign}(1-b/a))^n$ is positive, you are looking for lower bounds of the log-function for positive arguments, e.g. $\log (x) \ge 1 - 1/x$, which gives
$$
(a-b)^n \ge a^n \cdot ({\rm sign}(1-b/a))^n\cdot \exp (n (1 - 1/|1-b/a|))
$$
If $a^n\cdot({\rm sign}(1-b/a))^n$ is negative, you are looking for upper bounds of the log-function for positive arguments, e.g. $\log (x) \le x-1$.  Then you have
$$
(a-b)^n \ge a^n \cdot ({\rm sign}(1-b/a))^n \cdot \exp (n (|1-b/a| -1))
$$
The bounds on the log are sharp for $|1-b/a| =1$, and at some distance from this point, the deviations can become large. So with no more knowledge on  $|1-b/a|$, you can apply bounds on the log-function with more terms, e.g. $\log (x) \le x-1 - (x-1)^2/2 + (x-1)^3/3$ holds for all $x > 0$. If you have approximate knowledge on  the range of $|1-b/a|$, it is better to apply bounds  on the log-function which are centered to points which fall into that range.
A: For odd $n$, this is just impossible, because given any $c:=a^n$, $$d:=(\sqrt[n]c-b)^n$$ can take any value, by
$$b=\sqrt[n]c-\sqrt[n]d.$$
In other words, $a^n$ and $(a-b)^n$ are completely independent.
For even $n$, you have a mere $$(a-b)^n\ge0.$$
