How to prove inequality $ \vert A^{q}-1 \vert \le A^{\vert q \vert}-1$

Question

If q is rational and A is real $ > 1$, can you prove that
$$ \vert A^{q}-1 \vert \le A^{\vert q \vert}-1$$ ?

It should also be possible to extend the inequality to real exponents.

Answer 1

The case $q>0$ is obvious

If $q<0$ and $a>1$ then

$\begin{align} |A^q-1| \leq A^{|q|} -1 & \iff 1-A^q \leq A^{-q} -1 \\ &\iff A^q-(A^q)^2 \leq 1- A^{q} & \mathrm{because } \ A^q>0 \\ &\iff 0<(A^q)^2-2 A^q+1 \\ &\iff 0<(A^q-1)^2 & \mathrm{always \ true} \end{align}$

Answer 2

If $r\ge0$, then $A\gt1$ implies $A^r=A^{|r|}\gt1$, and thus

$$|A^r-1|=|A^{|r|}-1|=A^{|r|}-1$$

If $r\lt0$, then $A\gt1$ implies $0\lt A^r\lt1$, and thus

$$|A^r-1|=1-A^r=A^r(A^{-r}-1)=A^r(A^{|r|}-1)\lt A^{|r|}-1$$

This holds for all real $r$.