It is not difficult to transform your inequality to a simpler one by defining $x=c/w$ satisfying $0<x<1$, $c<w/2$ to have real solutions, and $w-2c>1$:
$$w^b\frac{(1-x)^{2 b+1}}{(1-2 x)^{b+1}}\geq 1$$
Plotting the rational function for several values of $0<b<1$, we see clearly that $x<0.5$:
GraphicsRow@(Plot[
Evaluate@
Table[(1 - x)^(1 + 2 b)/(1 - 2 x)^(
1 + b), {b, Range[1/10, 9/10, 1/10]}], {x, 0, #[[1]]},
PlotRange -> #[[2]], GridLines -> {None, {1}},
ImageSize -> 250] & /@ {{0.1, {0.75, 1.25}}, {0.49, {-1, 9}}})

We can inspect what happens when $x\rightarrow 0$ ($c\rightarrow 0$):
Limit[(1 - x)^(1 + 2 b)/(1 - 2 x)^(1 + b), x -> 0]
1
So it seems that the inequality is always fulfilled for $x=c/w<1/2$, and $w-2c>1$.
If $x=c/w$ satisfying $0<x<1$, $c<w/2$ to have real solutions, and $w-c<1$:
$$w^b\frac{(1-x)^{2 b+1}}{(1-2 x)^{b+1}}\leq 1$$
In this case as the rational function is monotonically increasing for different values of $b$, and $w\leq 1$, therefore, I presume some range for $c$ after solving numerically the above inequality given numerical values, and satisfying the initial conditions.
ApplySides
(introduced in v 11.3) and manually simplify $\endgroup$c<w/2
don't you? ( not that it leads to a reduction..) $\endgroup$c < w/2
$\endgroup$