# Bounding the ratio of the $\ell_1$-norms of two real-valued $n$-vectors as a linear combination of their $n$ element-wise ratios

Let $$a_1, a_2, \ldots a_n$$ and $$b_1, b_2, \ldots b_n$$ be two sequences of $$n\gg 1$$ real numbers such that, for all $$1\le i\le n$$, we have $$0. Let the ratio $$R$$ defined as follows:

$$R:=\frac{\sum_{1\le i\le n} a_i}{\sum_{1\le i\le n} b_i }~.$$

Question: How can we set values $$x_1, x_2, \ldots, x_n$$ as functions of the ratios $$r_i:=\frac{a_i}{b_i}$$ in such a way that $$R':=\sum_{1\le i\le n} (x_i\cdot r_i)$$ is a tight lower bound for $$R$$?

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 $$R=\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i} =\frac{\sum_{i=1}^n r_ib_i}{\sum_{i=1}^n b_i} \ge\frac{\sum_{i=1}^n r_* b_i}{\sum_{i=1}^n b_i}=r_*.$$