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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!

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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
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    $\begingroup$ Hint: (a+c)/(b+d) is avg (a,c)/avg (b,d) $\endgroup$ – Jacob Wakem Aug 17 '16 at 21:24
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    $\begingroup$ @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
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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: $$\frac{a_1+\dots+a_n}{b_1+\dots+b_n}\le \max\left(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n}\right).$$

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  • $\begingroup$ 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
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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:

$$(a+c)/(b+d)=(c/d)t+(a/b)(1-t)$$

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

$$d(bc-ad)/(b+d)=(bc-ad)t$$

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.

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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 $(d+b)$.

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  • $\begingroup$ 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
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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.

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