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Thread: Inequality Proof (not using Calculus)

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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.

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

    for $\displaystyle a > b > 0$
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    Anyone?
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    Sorry, my last proof was faulty. I'll have a look into it
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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.

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

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

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

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

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

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

    This is of the form

    $\displaystyle f(a) > f(b)$

    where

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

    Now all we need to establish is that

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

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

    Now, I had originally written:

    "Clearly, the numerator of $\displaystyle f(x)$ grows faster than the denominator, thus $\displaystyle f(x)$ 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 $\displaystyle f(x)$ 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 $\displaystyle g(u) = \log_x{u}$.

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

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

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

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

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

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

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

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

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

    $\displaystyle =x+1$

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

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

    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; Apr 17th 2010 at 12:24 AM.
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