Let $S$, $D$ and $r$ be non zero positive real numbers, where $0 < r < 1$, $D \geq S$, and let $k$ be a positive integer, where $k \geq 2$.
Find a general solution for $r$ given that: $$\frac{r^{k-1}(1-r)}{(1-r^{k})} \geq \frac{S}{D}.$$
I don't want to give anybody the idea that I haven't tried to solve this before putting it out there. I've tried shuffling around exponents, logarithmic identities and Mathematica :D but with no luck. It may be that I'm just missing something simple.
I'm not expecting a full solution but maybe some insight or a direction to take, but full and partial answers are all welcome.
Let $b = \frac{S}{D}$ Then we have $$\frac{r^{k-1}-1+1-r^k}{1 - r^k} \geq b $$ $$\frac{r^{k-1}-1}{1 - r^k} \geq b - 1 $$ If $c=b-1$ then it may be rewritten as $$c r^k + r^{k-1} - (c+1) \geq 0$$ which has no explicit solution for $k>4$, as it would be a polynomial of degree greater than $4$.