$x^a-(x-1)^aI work with math modelling and in one model I think it will be useful to solve the inequation
$$x^a-(x-1)^a<c^a,$$
with $0<c<1$ and $0<a<1$.
Do you have some hint? I need $x$ real, greater than $1$.
Thank you.
OBS.: (1) I need to iterate the solution for a lot of $a$'s (and a lot of $c$'s), so numerical solution is hard to work for many $c$′s. However, an aproximation for the solution would be OK!
(2) It would be ok if we consider $a\in\mathbb{Q}$.
 A: Divide by $x^a$,
$$1-(1-1/x)^a<(c/x)^a.$$
For simplicity write this as equation for $y=1/x$,
$$
1<(1-y)^a+(cy)^a
$$
Solvability
The right side at $y=1$ has the value $c^a$. The right side can be considered as a weighted average, so we can apply the mean value inequality for the concave function $f(u)=u^a$ and some weight $w^a$ for the second term
\begin{align}
(1-y)^{a}+w^a·(cy/w)^{a}
&\le(1+w^a) \left(\frac{(1-y)+w^{a-1}·cy}{1+w^a}\right)^{a}
\\
&=(1+w^a)^{1-a} \left(1+(cw^{a-1}-1)·y\right)^{a}
\end{align}
With $w=c^{1/(1-a)}$ this gives a maximum value of $(1+w^a)^{1-a}>1$ at
$$
1-y_0=cy_0/w\iff 1=(1+c^{-a/(1-a)})y_0
$$
You can take $x_0=1+c^{-a/(1-a)}$ as one solution for the inequality.
Getting closer to the boundary
At $y\lessapprox 1$ the first term will rise quickly from zero as $a<1$ gives a vertical tangent. So it makes sense to linearize this term by raising it to the power $1/a$,
$$
1-y>\left(1-(cy)^a\right)^{1/a},
$$
or use the power directly as new variable, $z=(1-y)^a$, $y=1-z^{1/a}$,
$$
z>1-(c(1-z^{1/a}))^a
$$
The right side is a smooth function around $y=1$, so it can be approximated by its tangent.
\begin{align}
1-y&\approx\left(1-c^a[1-a(1-y)]\right)^{1/a}
\\
&\approx\left(1-c^a\right)^{1/a}\left[1+\frac{c^a(1-y)}{1-c^a}\right]
\\
y_1&= 1-\frac{(1-c^a)^{1/a}}{1-c^a(1-c^a)^{1/a-1}}
\\
x_1&=\frac{1-c^a(1-c^a)^{1/a-1}}{1-c^a(1-c^a)^{1/a-1}-(1-c^a)^{1/a}}
=\frac{1-c^a(1-c^a)^{1/a-1}}{1-(1-c^a)^{1/a-1}}
=\frac{1-c^a}{1-(1-c^a)^{1/a-1}}+c^a
\end{align}
This gives a value closer to the intersection point, but usually slightly violating the desired inequality.
Empirically, this approximation is insufficient if $a\ge \min(0.9,0.75·c+0.4)$. For these one would have to start with approximations that use that now $(1-a)$ is small.

One can refine even that using the fixed-point iteration $y=g(y)=1-(1-(cy)^a)^{1/a}$. This converges slowly,  so apply an acceleration method like Aitkens delta-squared method. Also, there are cancellation problems for small values of $a$ and $y\approx 1$, as then the small number $(1-(cy)^a)$ gets even more reduced under the power $1/a$.
Or apply the Newton or secant method to $f(y)=(1-y)^a+(cy)^a$. Use over-relaxation to more quickly find a point on the other side of the root.
