$\frac{x}{y} \ge \frac{a_1}{b_1} \ge \frac{a_2}{b_2}$ and $b_1 \le b_2 \implies \frac{x+a_1}{y + b_1} \ge \frac{x+a_2}{y + b_2}$?

Question

Given $\frac{x}{y} \ge \frac{a_1}{b_1} \ge \frac{a_2}{b_2}$, where $x,y,a_i,b_i$ are positive numbers.

I would like to prove the following:

Claim: If $b_1 \le b_2$, then $\frac{x+a_1}{y + b_1} \ge \frac{x+a_2}{y + b_2}$.

Answer 1

We show $\frac{x+a_1}{y+b_1} - \frac{x+a_2}{y + b_2} \ge 0 $. \begin{align} &\frac{x+a_1}{y+b_1} - \frac{x+a_2}{y + b_2} = \frac{x(b_2 - b_1) + y(a_1 - a_2) + a_1 b_2 - a_2 b_1 }{(y+b_1)(y+b_2)} \nonumber\\ &\ge \frac{x(b_2 - b_1) + y(a_1 - a_2) }{(y+b_1)(y+b_2)} \end{align} We complete the proof showing $x(b_2 - b_1) \ge - y(a_1 - a_2)$. \begin{align} \frac{x}{y} + \frac{a_1 - a_2}{b_2 - b_1} \ge \frac{a_1}{b_1} + \frac{a_1 - a_2}{b_2 - b_1} = \frac{a_1 b_2 - a_2 b_1}{b_1 ( b_2 - b_1)} \ge 0 \end{align}