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If $\frac{a}{b}$ < $\frac{c}{d}$ then

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

I have been searching and trying and couldnt find a reliable proof.

One might do this like here which I think it is very wrong:

$\frac{a}{b}$ < $\frac{c}{d}$ then $a.d$ < $c.b$

Because $\frac{2}{-3}$ < $\frac{4}{5}$ doesn't mean $10 < -12$

Martin Sleziak
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Erfan Ahmadi
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    The proof you linked to mentioned that $a,b,c,d$ are positive real numbers – yoyostein Oct 28 '16 at 13:33
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    The middle fraction is called the mediant of the two others, and is between them in the order you state provided the first fraction is less than the second fraction. This assumes $a,b,c,d>0$ as would be usual for positive rationals. Try a google on "mediant"...[not "median", that's another concept] – coffeemath Oct 28 '16 at 13:39
  • Does it mean the inequality doesn't hold if they aren't all positive?@yoyostein – Erfan Ahmadi Oct 28 '16 at 13:51
  • What about the case $a=-2,b=1$ and $c=-3,d=-1$. – hamam_Abdallah Oct 28 '16 at 15:32
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    If searching "mediant" doesn't work, try "Farey fractions." The result you're trying to prove is a basic part of that theory. – B. Goddard Oct 28 '16 at 22:34
  • @Erfan I just put an answer giving a description of exactly when the desired inequality holds. It turns out to hold iff $b,d$ have the same signs, independently of signs on $a,c.$ – coffeemath Oct 29 '16 at 15:47
  • A few related posts: https://math.stackexchange.com/questions/205654/given-4-integers-a-b-c-d-0-does-fracab-fraccd-imply-fra https://math.stackexchange.com/questions/1761557/prove-fracacbd-lies-between-fracab-and-fraccd-for-posit https://math.stackexchange.com/questions/1891417/prove-that-fracab-fracacbd-frac-cd – Martin Sleziak Jan 27 '18 at 08:32
  • And you can find probably a few more related questions using Approach0. – Martin Sleziak Jan 27 '18 at 08:35

3 Answers3

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Here is a simple intuitive proof for the case when a, b, c and d are positive. Imagine a team that scores a points in its first b games and c points in the remaining d games. Its seasonal points per game is $\frac{a+c}{b+d}$, which can also be expressed as the weighted average of its performance in each part of the season.

$\frac{a+c}{b+d}$ = ($\frac{a}{b}$)($\frac{b}{b+d}$) + ($\frac{c}{d}$)($\frac{d}{b+d}$). This is a weighted average of $\frac{a}{b}$ and $\frac{c}{d}$, and so must between them.

user1153980
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In this answer we make no initial assumptions about the signs of $a,b,c,d$ and go for a characterization of when the desired inequality holds or does not hold.

Let $P=c/d-a/b=(bc-ad)/(bd),$ so that $P>0$ is the condition that $a/b<c/d$ which we are assuming. [Note that $bc-ad \neq 0$ since we are assuming $a/b<c/d.$] Also put $m=(a+c)/(b+d)$ which is the "middle term" of the desired inequality $a/b<m<c/d.$ Our claim is that this inequality holds if and only if $b,d$ have the same sign.

Now define $$\Delta_1=m-a/b=(bc-ad)/(b(b+d))=P\cdot (d/(b+d)),\\ \Delta_2=c/d-m=(bc-ad)/(d(b+d))=P \cdot (b/(b+d)).$$ Since $P>0,$ the desired inequality is thus equivalent to saying that $$\frac{d}{b+d}>0,\ \ \frac{b}{b+d}>0.\tag{1}$$ We are of course assuming none of $b,d,b+d$ are zero, in order that the three terms entering into the inequality all be defined.

Now first suppose $b,d$ have opposite signs. Then whatever the sign of $b+d$ be, one sees from (1) that the fractions mentioned in (1) cannot each be positive, so that in this case the desired inequality does not hold.

On the other hand, supposing $b,d$ have the same signs, it follows that both fractions in (1) are positive, so the desired inequality holds.

coffeemath
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Hint. As regards $\frac{a}{b} < \frac{a+c}{b+d}$, show that $$\frac{a+c}{b+d}-\frac{a}{b}=\frac{\frac{c}{d}-\frac{a}{b}}{\frac{b}{d}+1}.$$

Robert Z
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