An inequality involving fractions and a minimum

Question

It can be proved easily by contradiction that

if $a,b,c,d$ are positive numbers, then $$\frac{a+b}{c+d} \geq \min\Big\{ \frac{a}{c},\frac{b}{d}\Big\}.$$

I am not looking for a proof but rather for
1) a reference or book which contain this and similar inequalities;
2) information whether this inequality can be sharpened.

Thanks!

Answer 1

$$\frac{a+b}{c+d}$$ is the “mediant” of the fractions $\frac ac$ and $\frac bd$  – more precisely, the mediant of the ordered pairs $(a, c)$ and $(b, d)$. Your observation is the “mediant inequality”: If $a, b, c, d > 0$ then $$ \frac ac < \frac bd \quad \Longrightarrow \quad \frac ac < \frac{a+b}{c+d} < \frac bd \, . $$ This and more properties and applications of the mediant are described in Wikipedia: Mediant (mathematics).

The mediant can also be interpreted geometrically as the slope of the diagonal in a parallelogram, see here.

Answer 2

Let $\frac{b}{d}\geq\frac{a}{c}$.

Hence, we need to prove vthat $$\frac{a+b}{c+d}\geq\frac{a}{c}$$ or $$ac+bc\geq ac+ad$$or $$bc\geq ad,$$ which is our assuming.

By the same way we can check the case $\frac{b}{d}\leq\frac{a}{c}$.