This is from I. M. Gelfand's Algebra book.
Fractions $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are called
neighbor fractions if their difference $\displaystyle\frac{ad – bc}{bd}$ has numerator of $\pm 1$,
that is, $ad – bc = \pm 1$. Prove that(a) in this case neither fraction can be simplified (that is, neither has any
common factors in numerator and denominator);(b) if $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are neighbor
fractions, then $\displaystyle\frac{a+c}{b+d}$ is between them and is a neighbor fraction for
both $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$; moreover,(c) no fraction $\displaystyle\frac{e}{f}$ with positive integer $e$ and $f$ such
that $f < b + d$ is between $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$.
Parts (a) and (b) weren't too difficult, but I'm stuck on part (c). I've included (a) and (b) in case they're related to the solution to (c).
Best Answer
Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$
But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq \frac{1}{df}$. So $\frac{1}{bf} + \frac{1}{df} \leq \frac{1}{bd}$. Multiply both sides by $bdf$ and we get that $b+d\leq f$.