Tips on additional methods to solve inequality involving multiple variables So I want some feedback about my proof of the following implication:
Let $a,b,c,d\in\mathbb{Z}$ where $a,b,c,d>0$. Prove that if $\frac{a}{b}<\frac{c}{d}$, then 
$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
My attempt went as follows: 
If $\frac{a}{b}<\frac{c}{d}$, then 
$$ad<cb$$
Case 1: $\frac{a}{b}<\frac{a+c}{b+d}$
$$\frac{a}{b}=\frac{a}{b}\left(\frac{b+d}{b+d}\right)=\frac{ab+ad}{b(b+d)}$$
Since $ad<cb$
$$ab+ad<ab+cb\implies\frac{ab+ad}{b(b+d)}<\frac{ab+cb}{b(b+d)}\implies\frac{ab+ad}{b(b+d)}<\frac{b(a+c)}{b(b+d)}$$
$$\implies\frac{ab+ad}{b(b+d)}<\frac{a+c}{b+d}\implies\frac{a}{b}<\frac{a+c}{b+d}$$
Case 2: $\frac{a+c}{b+d}<\frac{c}{d}$
$$\frac{c}{d}=\frac{c}{d}\left(\frac{b+d}{b+d}\right)=\frac{cb+cd}{d(b+d)}$$
Since $ad<cb$
$$ad+cd<cb+cd\implies\frac{ad+cd}{d(b+d)}<\frac{cb+cd}{d(b+d)}\implies\frac{d(a+c)}{d(b+d)}<\frac{cb+cd}{d(b+d)}$$
$$\implies\frac{a+c}{b+d}<\frac{cb+cd}{d(b+d)}\implies\frac{a+c}{b+d}<\frac{c}{d}$$
So
$$\frac{a}{b}<\frac{a+c}{b+d}\land\frac{a+c}{b+d}<\frac{c}{d}\implies\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
So I'm quite sure this proof is a valid one, but I was wondering if there was a faster way of doing this. Maybe not requiring to break into two different cases but instead solving the compound inequality at once for both sides, or if there won't be much of a difference for both ways. Any tips would be appreciated.

I think your proof is right, but I like the following way.
$$\frac{a+c}{b+d}-\frac{a}{b}=\frac{bc-ad}{b(b+d)}>0$$ and
$$\frac{c}{d}-\frac{a+c}{b+d}=\frac{bc-ad}{d(b+d)}>0.$$
