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Real Analysis Proof

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Let y be a fixed real number satisfying 0<y[tex]\leq[/tex]1. Prove that (1+x)[tex]^{y}[/tex][tex]\leq[/tex]1+ x[tex]^{y}[/tex] for all x[tex]\geq[/tex]0.

    2. Relevant equations
    I'm not sure.

    3. The attempt at a solution
    The hint given with the problem states that the derivative of x[tex]^{y}[/tex] is yx[tex]^{y-1}[/tex]. My first thought is that I'm supposed to show that they are both strictly increasing, but I don't really know what that would help me with.

    I'm not really looking for an answer so much as a bit of direction.
  2. jcsd
  3. Dec 8, 2008 #2
    Are you sure you didn't reverse the inequality sign?
    Take f(x)=(1+x)y-1-xy and show that f' is positive for all positive x. This implies that f(x) is positive for all x because f(0)=0.
  4. Dec 8, 2008 #3
    Oh yes. I am quite certain that the inequality is right. Doing a few test cases shows that it is the correct inequality. So, basically, I end up with:

    -1<p[tex]\leq[/tex]0 Then letting p=|p|

    and f'(x)=y[tex]\frac{1}{x^{p}}[/tex]-[tex]\frac{1}{(1+x)^{p}}[/tex]. And since x[tex]\geq[/tex]0 for all x, and since x<x+1 for all x, x[tex]^{p}[/tex][tex]\leq[/tex](x+1)[tex]^{p}[/tex] and so [tex]\frac{1}{(x+1)^{p}}[/tex][tex]\leq[/tex][tex]\frac{1}{x^{p}}[/tex] giving that f'(x) is always positive and with f(0)=0, the intended result.

    I trust that this is in the right direction, and thank you very much for your response.
  5. Dec 8, 2008 #4
    I suppose I should do a separate case for p=0.
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