I would like to know if the following proposition holds or not.
For all $a, b, c, d$ such that $a, b, c, d$ are positive real numbers, IF $a < b$ and $a - c > b - d \geq 0$ THEN $\frac{c}{a} < \frac{d}{b}$ .
Could you kindly advise?
It is really appreciated if a proof or a counterexample is provided.
Best regards,
It's false. $a=1$ ; $b=2$ ; $c = 3$ ; $d=5$ is a counterexample.
We have $a<b$, $a-c=-2$ and $b-d = -3$ so $a-c>b-d$.
$\frac{c}{a}=3$ and $\frac{d}{b}=2.5$ so we don't have $\frac{c}{a} <\frac{d}{b}$
EDIT : With your new constraints, I believe this is true.
$a-c>b-d$ so $\frac{a}{c}-1> \frac{b-d}{c}$.
$d>c$ (because $a<b$ and $a-c>b-d$) so $\frac{1}{c} > \frac{1}{d}$ and since now we have $b-d\geq 0$, $\frac{b-d}{c} \geq \frac{b-d}{d} = \frac{b}{d}-1$.
So $\frac{a}{c}-1> \frac{b}{d}-1$ and $\frac{a}{c}> \frac{b}{d}$. So $\frac{c}{a}<\frac{d}{b}$.
