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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!

Michael Rozenberg
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max_zorn
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    since $1=\dfrac{\epsilon+n}{n+\epsilon}$ gives a min in $\dfrac{\epsilon}n$ it probably can be sharpened by considering $a\ge b$ and $c\ge d$, this has to be verified. – zwim Jan 19 '18 at 06:53
  • you want a reference but where did you take it from? – Guy Fsone Jan 19 '18 at 06:55
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    Compare https://math.stackexchange.com/questions/205654/given-4-integers-a-b-c-d-0-does-fracab-fraccd-imply-fra, where this is interpreted as the mediant, or as a slope in a parallelogram. – Martin R Jan 19 '18 at 10:08
  • @MartinR wow, thanks a lot! Please consider putting this down as the answer which it really is! – max_zorn Jan 19 '18 at 17:42

2 Answers2

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$$\frac{a+b}{c+d}$$ 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.

Martin R
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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 $\frac{b}{d}\leq\frac{a}{c}$.

Michael Rozenberg
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    Thanks very much, but as I wrote, I am really after a reference or whether this can be sharpened. – max_zorn Jan 19 '18 at 06:53