<s>[INST] 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 $10 < -12$
 [/INST] 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.
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