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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:

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


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 $R$?

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

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  • $\begingroup$ Thank you very much @Iosif Pinelis ! $\endgroup$ – Penelope Benenati Nov 3 '20 at 13:01

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