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 $0<a\_i \le b\_i\le 1$. Let the ratio $R$ defined as follows:


$$R:=\frac{\sum\_{1\le i\le n} a\_i}{\sum\_{1\le i\le n} b\_i }~.$$




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**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 $R$?



Bounding the ratio of the $\ell_1$-norms of two real-valued $n$-vectors as a linear combination of their $n$ element-wise ratios
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
$$R=\frac{\sum\_{i=1}^n a\_i}{\sum\_{i=1}^n b\_i}
=\frac{\sum\_{i=1}^n r\_ib\_i}{\sum\_{i=1}^n b\_i}
\ge\frac{\sum\_{i=1}^n r\_\* b\_i}{\sum\_{i=1}^n b\_i}=r\_\*.$$


