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 $\frac{b}{d}$.
[Edit: This question answers negatively the question "does $|a/c -x | > | x - b/d |$ imply $|b/d - mediant | > | mediant - x|$", which is not the one that was posed]
Your question appears to be "can the mediant never be more than $\frac 34$ of the way from $\frac ac$ to $\frac bd$ (in which case $x$ can be further from the mediant than from $\frac bd$ (necessarily on the opposite side) while still being closer to $\frac bd$ than to $\frac ac$. The answer is no (the mediant can be further off-center); take for instance $\frac ac=\frac11$. $\frac bd=\frac{99}{100}$, $\textit{mediant}=\frac{100}{101}$ and $x=\frac{149}{150}$. Then $|a/c-x|=\frac1{150}>|x-b/d|=\frac1{300}$ and $|b/d- mediant|=\frac1{10100} < | mediant - x|=\frac{49}{15150}$.
I have no idea why you think the property you stated would be true, as the second inequality will always fail for $x$ close enough to the mediant on the side of $b$. Take $\frac ac=\frac21$, $\frac bd=\frac11$, $\textit{mediant}=\frac32$ and $x=\frac43$ which is closer to $1$ than to $2$ and closer to $\frac32$ than to $1$.
$|\frac{a}{c}-x|>|x-\frac{b}{d}|$
$=>(\frac{a}{c}-x)^2>(x-\frac{b}{d})^2$
$=>(\frac{a}{c}-\frac{b}{d})(\frac{a}{c}+\frac{b}{d}-2x)>0$
$=>2x<\frac{a}{c}+\frac{b}{d}=>-x>\frac{-1}{2}(\frac{a}{c}+\frac{b}{d})$
$\frac{b}{d}-mediant=\frac{b}{d}-\frac{a+b}{c+d}=\frac{bc-ad}{d(c+d)}=\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})$
$mediant-x>\frac{a+b}{c+d} - \frac{1}{2}(\frac{a}{c}+\frac{b}{d})$
$2(mediant-x)>\frac{a+b}{c+d} - \frac{a}{c} + \frac{a+b}{c+d} - \frac{b}{d}$
$(mediant-x)>\frac{(bc-ad)(d-c)}{2cd(c+d)}=\frac{c-d}{2(c+d)}(\frac{a}{c}-\frac{b}{d})$
$|\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})|>$ 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.
