For each $i$, let $x_i:=r_*/(nr_i)$, where $r_*:=\min_j r_j$, so that $x_1,\dots,x_n$ are functions of the $r_i$'s. Then $$R'=\sum_{i=1}^n x_i r_i=r_*\le R,\tag{1}$$ so that $R'$ is a lower bound on $R$. This bound is tight, since $R'=r_*=R$ is the $r_i$'s are the same for all $i$.

The inequality in (1) holds because $$R=\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i} =\frac{\sum_{i=1}^n r_ib_i}{\sum_{i=1}^n b_i} \ge\frac{\sum_{i=1}^n r_* b_i}{\sum_{i=1}^n b_i}=r_*.$$

For each $i$, let $x_i:=r_*/(nr_i)$, where $r_*:=\min_j r_j$, so that $x_1,\dots,x_n$ are functions of the $r_i$'s. Then $$R'=\sum_{i=1}^n x_i r_i=r_*\le R,\tag{1}$$ so that $R'$ is a lower bound on $R$. This bound is tight, since $R'=r_*=R$ if the $r_i$'s are the same for all $i$.

The inequality in (1) holds because $$R=\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i} =\frac{\sum_{i=1}^n r_ib_i}{\sum_{i=1}^n b_i} \ge\frac{\sum_{i=1}^n r_* b_i}{\sum_{i=1}^n b_i}=r_*.$$