Properties of Mediants If $\frac{a}{c} > \frac{b}{d}$, then the mediant of these two fractions is defined as $\frac{a+b}{c+d}$ and can be shown to lie striclty between the two fractions. 
My question is can you prove the following property of mediants: if $|\frac{a}{c} - x| > |x - \frac{b}{d}|$ then $|b/d - mediant| < |mediant - x|$ for any $x$ that lies strictly between $\frac{a}{c}$ and $\frac{b}{d}$. 

$|\frac{a}{c}-x|>|x-\frac{b}{d}|$
$=>(\frac{a}{c}-x)^2>(x-\frac{b}{d})^2$
$=>(\frac{a}{c}-\frac{b}{d})(\frac{a}{c}+\frac{b}{d}-2x)>0$
$=>2x<\frac{a}{c}+\frac{b}{d}=>-x>\frac{-1}{2}(\frac{a}{c}+\frac{b}{d})$
$\frac{b}{d}-mediant=\frac{b}{d}-\frac{a+b}{c+d}=\frac{bc-ad}{d(c+d)}=\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})$
$mediant-x>\frac{a+b}{c+d} - \frac{1}{2}(\frac{a}{c}+\frac{b}{d})$
$2(mediant-x)>\frac{a+b}{c+d} - \frac{a}{c} + \frac{a+b}{c+d} - \frac{b}{d}$
$(mediant-x)>\frac{(bc-ad)(d-c)}{2cd(c+d)}=\frac{c-d}{2(c+d)}(\frac{a}{c}-\frac{b}{d})$ 
$|\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})|>$ will be greater than |$\frac{c-d}{2(c+d)}(\frac{a}{c}-\frac{b}{d})|$
iff  $|\frac{-c}{c+d}|>|\frac{c-d}{2(c+d)}|$
if  $|-2c|>|c-d|$ when c+d>0  or if $|-2c|<|c-d|$ when c+d<0
if  $2>|1-\frac{d}{c}|$ when c+d>0  or if $2<|1-\frac{d}{c}|$ when c+d<0
In the 1st case, $-2<1-\frac{d}{c}<2=>-3<-\frac{d}{c}<1=>-1<\frac{d}{c}<3$
c+d>0 which is true and d<3c
In the 2nd case, $1-\frac{d}{c}>2$ or $1-\frac{d}{c}<-2$
or $\frac{d}{c}<-1$ or $\frac{d}{c}>3$
or c+d<0 which is true or d>3c
So the conditions are $(c+d>0\ and\ d<3c)$ or $(c+d<0\ and\ d>3c)$ to make the inequality valid.
