Given 4 integers, $a, b, c, d > 0$, does $\frac{a}{b} < \frac{c}{d}$ imply $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$?

We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$?

Yes, the inequality holds. One standard approach to proving that $x\lt y\,$ is to show that $y-x\gt 0$.

Apply this to $x=\dfrac{a}{b}$ and $y=\dfrac{a+c}{b+d}$.

The difference is $\dfrac{a+c}{b+d}-\dfrac{a}{b}$, which simplifies to $\dfrac{bc-ad}{(b+d)b}$. But $bc\gt ad$ follows from our initial inequality.

The same method works for showing that $\dfrac{a+c}{b+d}\lt \dfrac{c}{d}$.

Hint $\ $ The middle term, known as the mediant, is the slope of the diagonal of the parallelogram with sides being the vectors $(b,a),\ (d,c).\:$ Clearly the slope of the diagonal lies between the slopes of the sides.