To put it briefly, my question is : suppose $a>b$ and $n>0$, how does $\frac {1} {a^n}$ compare to $\frac {1} {b^n}$ ?
I have considered various cases without arriving at finding a general rule.
In view of deriving the order relation between $\frac {1}{a^n}$ and $\frac {1}{b^n}$ in each case, I use this rule : let a given ordering relation ( greater than/ less than) hold between $N$ and $M$, then
if $N$ and $M$ have the same sign, their ( multiplicative) inverses have the reverse order
if $N$ and $M$ have opposite signs, then, the ( multiplicative ) inverses preserve the order.
I apply this rule to the $n$th power of $a$ and of $b$, previously ordered in each case.
My "strategy" was as follows: (1) first determining the order relation of the $n$th powers, and then (1) deriving from this the order relation of the inverses of the $n$th powers. But finally, what I end up with is a mess.
I managed to find a sort of rule for the $n$th powers, but not for their inverses. The rule for $n$th powers was as follows :
"In case a> b , and n > 0 , then $n$th-powers conserve the order, that is , $a^n > b^n$, except when $n$ is even and either (1) $a$ and $b$ are both negative , or (2) $a$ and $b$ have different signs and $a$ is smaller than $b$ in absolute value."
If there a way to find a general rule for the cases distinguished below.

If $n>0$ is an integer, this is $n\in\mathbb{Z}^+$, and $a>b$ are any two real numbers, then we have even nine different cases:
What happens then?
So be careful because