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!
 A: Suppose we have positive $a,b,c,d$ with $\frac{a}{b}\ge \frac{c}{d}$. Then multiplying through by $bd$ we get $ad\ge bc$. Adding $ab$ to both sides we get $a(b+d)\ge b(a+c)$. Dividing by $b(b+d)$, we get $\frac{a}{b}\ge\frac{a+c}{b+d}$.
Similarly, add $cd$ to both sides of $ad\ge bc$ to get $d(a+c)\ge c(b+d)$. Dividing by $d(b+d)$ we get $\frac{a+c}{b+d}\ge \frac{c}{d}$.
A: Assuming $a,b,c,d\gt0$, we have
$$\begin{align}
{a\over b}\lt{a+c\over b+d}\lt{c\over d}&\iff a(b+d)\lt b(a+c)\quad\text{and}\quad(a+c)d\lt(b+d)c\\
&\iff ad\lt bc\quad\text{and}\quad ad\lt bc\\
&\iff ad\lt bc\\
&\iff {a\over b}\lt{c\over d}
\end{align}$$
A: Here is another way that uses calculus:
Let $\phi(t) = {(1-t)a+t c \over (1-t)b + t d}$ and note that
$\phi(0) = {a \over b}, \phi(1) = {c \over d}$. Furthermore,
$\phi'(t) = {bc-ad \over ((1-t)b + t d )^2}$. Since $ad < bc$, we see that
$\phi$ is increasing, and so
$\phi(0) \le \phi({1 \over 2}) \le \phi(1)$. Since
$\phi({1 \over 2}) = {a+c \over b+d}$, we have the desired result.
A: 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 $$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}.$$
A: Here's one way to look at it: 
You're taking a class.  Suppose you get $a$ points out of $b$ possible on Quiz 1, and $c$ points out of $d$ possible on Quiz 2.  
Your overall points are $a+c$ out of $b+d$ possible.  And your overall percentage should be between your lower quiz score and your higher quiz score.
A: Hint $\ $ The mediant $(a\!+\!b)/(c\!+\!d)\,$ is the slope of the diagonal of the parallelogram with vector sides $(b,a),\ (d,c).\:$ But the slope of the diagonal lies between the slopes of the sides.
