If $0
If $0<a<b$ and $0<c<d$ then $\frac{c+a}{d+a} <\frac{c+b}{d+b}.$
I get to $$d+a<d+b \Longrightarrow \frac{1}{d+b} < \frac{1}{d+a}$$ but that inequality seems opposite of what I am trying to prove. Any advice is appreciated.

$$ \dfrac{c+a}{d+a} =\dfrac{d+a+c-d}{d+a}  = 1 + \dfrac{c-d}{d+a}$$
$$ \dfrac{c+b}{d+b} =\dfrac{d+b+c-d}{d+b}  = 1 + \dfrac{c-d}{d+b}$$
You have already proved that
$$ \frac{1}{d+b} < \frac{1}{d+a}$$
As $c-d<0$,
$$ \frac{c-d}{d+b} > \frac{c-d}{d+a}$$
$$ \implies1+ \frac{c-d}{d+b} > 1+\frac{c-d}{d+a}$$
$$ \implies \dfrac{c+b}{d+b} > \dfrac{c+a}{d+a}$$
