The $r_k$'s are terms of a sequence of positive integers satisfying $\sup r_k=\infty$. I am looking for a solution $F_k$ to
$$F_{k}\ge 2r_{k+1} F_{k+1}-F_{k+1}\ (k\ge1);\ F_1=1 \ (*)$$
(I want $F_{k}$ to be the smallest possible solution to $(*)$). Equality if possible.
Take $F_{k}=q^{k}, q\not=0.$ Then we have $$q^{k}-2r_{k+1}\cdot q^{k+1}+q^{k+1}\ge 0,$$ $$q^{k}(1-2r_{k+1}\cdot q+q)\ge 0$$ Therefore, since $q\not =0$,
$$1-2r_{k+1} q+q\ge0 $$ Solving for $q$ gives $$q\le \frac{1}{2r_{k+1}-1}$$
Therefore, $F_{k}\le (\frac{1}{2r_{k+1}-1})^{k}.$
Since $(*)$ isn't linear and we're dealing with an inequality, how do I know if I have a general enough solution of $(*)$?
Your inequality rearranges to $$\frac{F_k}{2r_{k+1}-1}\ge F_{k+1}, k\ge 1$$
Hence $$F_2\le F_1 \left(\frac{1}{2r_2-1}\right)$$ and $$F_3\le F_2\left(\frac{1}{2r_3-1}\right)\le F_1 \left(\frac{1}{2r_3-1}\right)\left(\frac{1}{2r_2-1}\right)$$
Hence you can prove by induction that $$F_n\le F_1\prod_{i=2}^n\left(\frac{1}{2r_i-1}\right)$$
(and equality can be achieved)