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Is the question too general to answer? I'm talking about when the former expression will be greater (or smaller) than the latter?

Here's an extension:

Compare expressions $$p\frac{a}{b}+q\frac{c}{d}+r\frac{e}{f}......$$

And

$$(p+q+r+.....)(\frac{a+c+e+....}{b+d+f+...})$$

For positive $a,b,c,d...$ and $p,q,r,s.....$. This is of course assuming that the first sequence of variables doesn't have variables having the same names has those in the second sequence (say, we start naming them by greek letters once we reach $p$).

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  • $\begingroup$ It is too general. Why don't you play around yourself for a while with just two summands? Do some numerical tests. Fix some of the variables and see what happens as the other(s) vary. $\endgroup$ – Ethan Bolker May 5 at 14:12
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    $\begingroup$ Hint: math.stackexchange.com/questions/751155/… Can you extend the result by induction from a pre-sorted list of fractions ? $\endgroup$ – zwim May 5 at 14:13
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If the variables are all positive, $\frac{a}{b}+\frac{c}{d}+… \gt \frac{a+c+…}{b+d+…}$. Break up the right side as $\frac a{b+d+\ldots}+\frac c{b+d+\ldots}+\ldots$ and notice that each term on the right has a matching one on the left, but the one on the right has greater denominator.

Your extension can go either way, because the product has many cross terms.

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  • $\begingroup$ Thanks. That was too easy. But I had an extension of the question in mind which I've put in the edit. $\endgroup$ – Ryder Rude May 5 at 14:18
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    $\begingroup$ I commented on the extension. It should be easy to find versions, even with two of each kind of variable, that make the inequality go either way. In one case you want the cross terms large, in the other small. $\endgroup$ – Ross Millikan May 5 at 14:21

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