Assume we have a finite collection of real values $\{a_1,\dots,a_n\}$ each between 0 and 1, and their sum less than 1. The same for the set $b$.
The inequality
$$\frac{a_i}{x} > \frac{b_i}{y} \quad (*)$$
must hold for all $i$, with both $x$ and $y$ also between 0 and 1. Is there a test I can do on the set $a$ and $b$ to guarantee (*), without any knowledge of $x$ or $y$?
An incorrect thought was to divide two different labeled inequalities producing
$$\frac{a_i}{a_j} > \frac{b_i}{b_j} $$
which must hold for ordered ${i,j}$ ($a_i > a_j$), but that too also didn't work.
Assume that $a_i, b_i, y , x \in (0,1)$.
If $\max_i\frac{b_i}{ya_i}\le 1$, then we have (*) holds since $\frac1x>1\ge\max_i\frac{b_i}{ya_i}$ would implies that for all $i$, $$\frac{a_i}x>\frac{b_i}{y}$$
However, if $\max_i\frac{b_i}{ya_i} > 1$, then we can't conclude as it is possible to have a small $x$ such that the inequality holds but at the same time it is possible to have $x$ that is not small enough.