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Math Help - Inequality Proof (not using Calculus)

  1. #1
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    Inequality Proof (not using Calculus)

    Hey guys,

    Just wondering would you be able to help me with this question. I know of a proof using calculus, but I think there may be another method.

    \left( {\frac{a+1}{b+1}}\right) ^{b+1} > \left( \frac{a}{b} \right) ^{b}

    for a > b > 0
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  2. #2
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    Anyone?
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  3. #3
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    Sorry, my last proof was faulty. I'll have a look into it
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  4. #4
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    Quote Originally Posted by alan4cult View Post
    Hey guys,

    Just wondering would you be able to help me with this question. I know of a proof using calculus, but I think there may be another method.

    \left( {\frac{a+1}{b+1}}\right) ^{b+1} > \left( \frac{a}{b} \right) ^{b}

    for a > b > 0
    (Edited) Hmm, I thought I had come up with a nifty proof, but seems now that I wasn't able to avoid calculus.

    \left( {\frac{a+1}{b+1}}\right) ^{b+1} > \left( \frac{a}{b} \right) ^{b}

    \iff {\frac{(a+1)^{b+1}}{(b+1)^{b+1}}}>  \frac{a^b}{b^b}

    \iff (a+1)^{b+1}b^b > (b+1)^{b+1}a^b

    \iff \frac{(a+1)^{b+1}}{a^b} > \frac{(b+1)^{b+1}}{b^b}

    This is of the form

    f(a) > f(b)

    where

    f(x) = \frac{(x+1)^{b+1}}{x^b}

    Now all we need to establish is that

    a > b \iff f(a) > f(b)

    In other words, we need to show that f(x) is increasing.

    Now, I had originally written:

    "Clearly, the numerator of f(x) grows faster than the denominator, thus f(x) is increasing and the proof is concluded."

    But I don't think that's quite good enough, and now I'm not sure how to prove rigorously that f(x) is increasing without derivatives... maybe someone else can see a way?

    EDIT: I'm not confident about the following method, but perhaps it is valid:

    Let g(u) = \log_x{u}.

    Since g(u) preserves order (is monotonic), applying g(u) to both sides of an equation will also preserve order. So:

    f(x) = \frac{(x+1)^{b+1}}{x^b}

    \iff g(f(x)) = g\left(\left(\frac{x+1}{x}\right)^b (x+1)\right)

    = (b+1)\log_x(x+1)-b\log_x{x}

    = (b+1)\log_x(x+1)-b

    Let h(x)=\log_x(x+1). Then it is sufficient to show that h(x) is increasing, in order to prove that f(x) is increasing.

    Now let k(u) = x^u. This is again order preserving, and we have:

    h(x)=\log_x(x+1)

    \iff k(h(x))=k\left(\log_x(x+1)\right)

    =x+1

    It is clear that x+1 is increasing, therefore f(x) is increasing and the proof is concluded.

    I'm bothered by the result because, when I took derivatives, I found that f'(x)=0 when x=b, and nowhere in the above steps did I utilize the restriction x \geq b.

    Ah well, I've edited this post about a million times, I'll leave it in the hands of other, smarter people.
    Last edited by undefined; April 17th 2010 at 12:24 AM.
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