Prove: $\frac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$ (for positive $a$, $b$, $c$, $d$) I am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.
Would be grateful for any suggestions!

Assume that $b,d >0$. Note that $$\frac{a+c}{b+d} = \frac{b}{b+d}\frac{a}{b} +\frac{d}{b+d}\frac{c}{d}.$$ Remark that $0<\frac{b}{b+d}<1$ and the same for $\frac{d}{b+d}$. Hence you have written $\frac{a+b}{c+d}$ as a convex combination of $\frac{a}{b}$ and $\frac{c}{d}$ so you get $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}.$$
