Mediant Inequality Proof: $\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$ If $\frac{a}{b}$ < $\frac{c}{d}$ then 
$$\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$$ 
I have been searching and trying and couldnt find a reliable proof. 
One might do this like here which I think it is very wrong:
$\frac{a}{b}$ < $\frac{c}{d}$ then $a.d$ < $c.b$
Because $\frac{2}{-3}$ < $\frac{4}{5}$ doesn't mean $10 < -12$
 A: Here is a simple intuitive proof for the case when a, b, c and d are positive. Imagine a team that scores a points in its first b games and c points in the remaining d games.  Its seasonal points per game is $\frac{a+c}{b+d}$, which can also be expressed as the weighted average of its performance in each part of the season.
$\frac{a+c}{b+d}$ = ($\frac{a}{b}$)($\frac{b}{b+d}$) + ($\frac{c}{d}$)($\frac{d}{b+d}$). This is a weighted average of $\frac{a}{b}$ and $\frac{c}{d}$, and so must between them.
A: In this answer we make no initial assumptions about the signs of $a,b,c,d$ and go for a characterization of when the desired inequality holds or does not hold.
Let $P=c/d-a/b=(bc-ad)/(bd),$ so that $P>0$ is the condition that $a/b<c/d$ which we are assuming. [Note that $bc-ad \neq 0$ since we are assuming $a/b<c/d.$] Also put $m=(a+c)/(b+d)$ which is the "middle term" of the desired inequality $a/b<m<c/d.$ Our claim is that this inequality holds if and only if $b,d$ have the same sign.
Now define 
$$\Delta_1=m-a/b=(bc-ad)/(b(b+d))=P\cdot (d/(b+d)),\\ 
\Delta_2=c/d-m=(bc-ad)/(d(b+d))=P \cdot (b/(b+d)).$$
Since $P>0,$ the desired inequality is thus equivalent to saying that
$$\frac{d}{b+d}>0,\ \ \frac{b}{b+d}>0.\tag{1}$$
We are of course assuming none of $b,d,b+d$ are zero, in order that the three terms entering into the inequality all be defined.
Now first suppose $b,d$ have opposite signs. Then whatever the sign of $b+d$ be, one sees from (1) that the fractions mentioned in (1) cannot each be positive, so that in this case the desired inequality does not hold.
On the other hand, supposing $b,d$ have the same signs, it follows that both fractions in (1) are positive, so the desired inequality holds.
A: Hint. As regards $\frac{a}{b} < \frac{a+c}{b+d}$, show that
$$\frac{a+c}{b+d}-\frac{a}{b}=\frac{\frac{c}{d}-\frac{a}{b}}{\frac{b}{d}+1}.$$
