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Math Help - Inequality proof.

  1. #1
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    Exclamation Inequality proof.

    I need to mathematically prove that:

    (a + b)/(c + d) lies between (a / c) and (b / d).

    a,b,c,d are all real and >0

    Any ideas?

    (also, if necessary you can assume that b>a and d>c but preferably not)
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  2. #2
    Junior Member
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    Oct 2009
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    We don't need to assume b\geq a (or the other way), but we do have to assume either ad \leq bc or the opposite*. It's obviously one of them (unless it's both) so there's no danger in doing this; also notice that the roles of a and d can be switched with those of b and c respectively in the middle of the inequality, so what we're doing is all kosher.

    So the question is: Prove \frac{a}{c} \leq \frac{a+b}{c+d} \leq \frac{b}{d}

    Start with the assumption ad < bc (When I write < I mean less than or equal to, but I don't want to format this entire answer).

    First, add ac to both sides:
    ad+ac < bc+ac
    a(d+c) < c(a+b)

    a/c < (a+b)/(c+d). Voila, the left inequality

    Then go back to the assumption but add bd to both sides:
    ad+bd < bc+bd
    d(a+b) < b(c+d)

    (a+b)/(c+d) < b/d. And there's the right one.

    [ Note: when I did my scratchwork, I did it by assuming what I wanted to prove and tried to get something I knew to be true; I didn't just pull "+bd" out of nowhere. But that's not actually a proof - you're not allowed to assume what you want to be true and then use it to show 0=0, the logic can fail ]

    * This may seem like a bizarre thing to assume, but if I wrote it like this: "We must assume either a/c < b/d OR b/d < a/c," now it makes more sense, doesn't it? I chose the non-fraction form because it was easier for the computation in the proof, also it's easier to see how the a/b // d/c pairs can be switched around.
    Last edited by mr fantastic; May 16th 2011 at 07:23 PM.
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