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Inequality proof homework

  1. May 1, 2007 #1
    1. The problem statement, all variables and given/known data

    If we know that [tex](\frac{a - 1}{1 + a})^{n + 1} \geq \displaystyle\prod_{i=0}^n\frac{x_i - 1}{1+x_i}[/tex] is the inequality
    [tex] a^{n+1} \geq \displaystyle\prod_{i=0}^n\ x_i [/tex] true? Prove your answer.

    2. Relevant equations

    Not sure

    3. The attempt at a solution

    I tried induction:

    The base n = 0 works.

    Assume it works for n -1

    Proving it works for n:

    [tex] a^{n +1} = aa^n \geq a\displaystyle\prod_{i=0}^{n - 1} x_i

    = \frac{a}{x_n}\displaystyle\prod_{i=0}^{n} x_i [/tex].

    Now it would be great if I could assume that if it works for n = 0 then
    [tex] a \geq x_0 [/tex] and therefore [tex] a \geq x [/tex] for all n since I can allways permute the highest of the x and set it as [tex] x_0[/tex]. If this is true, then I would get the result immediately. But I don't really know if I could do this. Any help is appreciated. I am very interested to see if the inequality could be proven without induction. Thanks for any comments.
     
    Last edited: May 1, 2007
  2. jcsd
  3. May 1, 2007 #2
    wow my latex sucks. sorry for that. Im trying to fix some
     
  4. May 1, 2007 #3
    Can it be proven here that [tex] a \geq x [/tex] for all n, or should I just say that the inequality is true iff [tex] a \geq x [/tex] for all n?
     
  5. May 1, 2007 #4
    Oh I got it now. I didn't realize that

    [tex](\frac{a - 1}{a + 1})^{n + 1}

    = \displaystyle\prod_{i=0}^n\ (\frac{a - 1}{a + 1})

    \geq \displaystyle\prod_{i=0}^n\ \frac{x - 1}{1 + x} [/tex].

    But since

    [tex] f(x) = \frac{x - 1}{1 + x}[/tex]

    is an increasing function, then a must be greater than x for all n. Is this reasoning correct? I am suspicious of that. I need some confirmation. Thanks
     
    Last edited: May 1, 2007
  6. May 1, 2007 #5
    Wait, no I don't think that's true. Man!
     
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