I need help in proving (or disproving) the following assumption:

$$\frac{a + c}{b + d} \leq \max(\frac{a}{b},\frac{c}{d})$$

where $a,b,c,d \geq 0$ are positive integers. Both fractions $\frac{a}{b}$ and $\frac{c}{d}$ are between 0 and 1 and therefore the conditions $a \leq b$ and $c \leq d$ hold.

Any help or ideas are appreciated, thank you!


marked as duplicate by Dragonemperor42, heropup, Shailesh, Chill2Macht, user91500 Aug 23 '16 at 4:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Are $a, b, c, d$ all positive? Because otherwise $(7, 8, -5, -7)$ is a counterexample. $\endgroup$ – Brian Tung Aug 17 '16 at 21:12
  • $\begingroup$ For completeness, you could add that $a,b,c,d$ are positive integers (I assume), and anything you have tried already that hasn't worked. $\endgroup$ – 6005 Aug 17 '16 at 21:13
  • $\begingroup$ @6005 Yes, $a,b,c$ and $d$ are positive integers. I edited the question to make this more clear. $\endgroup$ – Michael Rapp Aug 17 '16 at 21:17
  • 2
    $\begingroup$ Hint: (a+c)/(b+d) is avg (a,c)/avg (b,d) $\endgroup$ – Jacob Wakem Aug 17 '16 at 21:24
  • 2
    $\begingroup$ @George V. Williams I wouldn't say it is a "duplicate" but an immediate consequence of the good reference you gave. $\endgroup$ – Jean Marie Aug 17 '16 at 21:29

Assume without loss of generality that $\dfrac{c}{d}\geq\dfrac{a}{b}$, i.e., $bc\ge ad$.

You want to show that $\dfrac{c}{d}-\dfrac{a+c}{b+d}\ge 0$. This means $\dfrac{bc-ad}{d(b+d)}\ge 0$. But this is true from the above.

The inequality can be generalized to more variables: $$\frac{a_1+\dots+a_n}{b_1+\dots+b_n}\le \max\left(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n}\right).$$

  • $\begingroup$ I think this is the way to go. I found an identical, but more detailed proof here: joelreyesnoche.files.wordpress.com/2012/11/mediant.pdf. But where does the condition $bc \geq ad$ come from? $\endgroup$ – Michael Rapp Aug 17 '16 at 21:53
  • $\begingroup$ It comes from the condition $\frac{c}{d}\geq\frac{a}{b}$, which can be assumed without loss of generality because of the symmetry between $\frac{c}{d}$ and $\frac{a}{b}$. Otherwise, if you prefer, you can consider the two cases $\frac{c}{d}\geq\frac{a}{b}$ and $\frac{c}{d}\leq\frac{a}{b}$. $\endgroup$ – pi66 Aug 17 '16 at 21:56
  • $\begingroup$ Yes, I understand that $\frac{a}{d} \geq$ can be assumed without loss of generality, but it isn't very intuitive to me that $bc \geq ad$ follows from it. I am not a mathematician, maybe this is common knowledge... $\endgroup$ – Michael Rapp Aug 17 '16 at 22:00
  • $\begingroup$ $bc\geq ad$ just follows from $\frac{c}{d}\geq\frac{a}{b}$ by cross-multiplication. Is it clear now? :) $\endgroup$ – pi66 Aug 17 '16 at 22:02
  • $\begingroup$ Alright, I got it. It is in fact intuitive :-D Thank you very much for your help! I will mark this as the correct answer. $\endgroup$ – Michael Rapp Aug 17 '16 at 22:08

We wish to show that the inequality holds, with equality only when $a/b=c/d$

Consider the function $f(t)=(c/d)t+(a/b)(1-t)$. This is a linear function equal to $a/b$ when $t=0$ and $c/d$ when $t=1$. Moreover, $f(t)$ is between these two quantities if and only if $0\leq t\leq 1$. Therefore, if we can show that $f(t)=(a+c)/(b+d)$ for some $t$ between $0$ and $1$, we will be done.

Let us solve:


Multiplying by $b*d$, expanding out, and rearranging terms, we get


Assuming that $bc=ad$ is nonzero (which happens when $a/b$ and $c/d$ are not equal), we can divide out to get $t=d/(b+d)$. Assuming that $d>0$ and $b>0$, or that $b<0$ and $d<0$ we have $t$ is between $0$ and $1$, and we are done. However, if the signs of $b$ and $d$ are different, then $t$ will be outside the range, and the inequality will not hold.


Holders inequality (the $p=\infty$,$q=1$ case) has: $$a+c =\frac{a}{b} b + \frac{c}{d} d \leq \max\{\frac{a}{b},\frac{c}{d} \}b+\max\{\frac{a}{b},\frac{c}{d} \}d = \max\{\frac{a}{b},\frac{c}{d} \}\left(b+d\right)$$ Nonnegativity is needed in the step with the inequality and in dividing by $(d+b)$.

  • $\begingroup$ Holder's inequality? Where did that come from? Your solution just used simple algebra. $\endgroup$ – marty cohen Aug 17 '16 at 22:13
  • $\begingroup$ It's the $p=\infty$, $q=1$ case. $\endgroup$ – abnry Aug 17 '16 at 22:47
  • $\begingroup$ All you used was $x \le \max(x, y)$. Very unprofound. $\endgroup$ – marty cohen Aug 17 '16 at 23:36
  • $\begingroup$ I'll try to knock your socks off with profundity next time. =P $\endgroup$ – abnry Aug 18 '16 at 0:55

Without loss of generality, assume a/b ≤ c/d. Then ad ≤ bc and ad +cd ≤ bc +cd so that (a+c)d ≤ (b+d)c. It follows that (a+c)/(b+d) ≤ c/d. There is no need to assume that the fractions are between 0 and 1. A similar argument shows that (a+c)/(b+d) lies between the min and the max of a/b, c/d. Google "Farey addition" and you will see that there is much more to adding fractions the way you always wanted to than you would guess.


Not the answer you're looking for? Browse other questions tagged or ask your own question.