Given two fractions $\frac{h}{k}$ and $\frac{h^{'}}{k^{'}}$ both in reduced form. I am unable to find a case when $\frac{h+h^{'}}{k+k^{'}}$ does not lie in the interval $\big[ \frac{h}{k},\frac{h^{'}}{k^{'}} \big]$. Is there such a case ?
PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.
On
Let's prove that, for positive $a,b,c,d$, with $\frac{a}{b}\le\frac{c}{d}$, it holds $$ \frac{a}{b}\le\frac{a+c}{b+d}\le\frac{c}{d} $$ The inequality on the left is equivalent to $$ ab+ad\le ab+bc $$ that is, $ad\le bc$, which is true.
The inequality on the right is equivalent to $$ ad+cd\le bc+cd $$ that is, $ad\le bc$, which is true.
Note that, if $a/b<c/d$, the two inferred inequalities are strict too. The hypothesis about the fractions being in reduced form is redundant.
Let $\dfrac hk=a$ and $\dfrac {h'}{k'}=b$.
Then, we have $h=ak$ and $h'=ak'$.
Then, $\dfrac{h+h'}{k+k'}=\dfrac{ak+bk'}{k+k'}$.
WLOG, let $a<b$.
$\dfrac{ak+bk'}{k+k'}>\dfrac{ak+ak'}{k+k'}=a$
$\dfrac{ak+bk'}{k+k'}<\dfrac{bk+bk'}{k+k'}=b$
$a<\dfrac{ak+bk'}{k+k'}<b$
$\dfrac hk<\dfrac{h+h'}{k+k'}<\dfrac{h'}{k'}$
The case where $a=b$ and where $a>b$ is left to the reader as an exercise.