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Simple questions about some inequalitites

  1. May 7, 2010 #1
    If [itex]a[/itex] and [itex]b[/itex] are positive and [itex]a < b[/itex], do we have

    (0 < x < 1) \Rightarrow \frac{1}{x^b} > \frac{1}{x^a}


    (1 < x < \infty) \Rightarrow \frac{1}{x^a} > \frac{1}{x^b}

    Last edited: May 7, 2010
  2. jcsd
  3. May 7, 2010 #2


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    I think you should be able to prove these yourself. Here's the first one:

    Fact: Let [itex]0<x<1[/itex]. Then for all [itex]\alpha>0[/itex] we have [itex]0<x^\alpha<1[/itex] and (hence) [itex]\frac{1}{x^\alpha}>1[/itex].

    In particular:
    * [itex]0<x^a<1[/itex] and [itex]0<x^b<1[/itex];
    * [tex]\frac{x^a}{x^b}=\frac{1}{x^{b-a}}>1,[/tex] hence [itex]x^a>x^b[/itex].

    Together: [itex]0<x^b<x^a<1[/itex]. Conclusion:
  4. May 7, 2010 #3
    Thanks for your help, Landau. This is one of those questions that I answered for myself when I was typing it up, but I thought I'd go ahead and post it anyway to make sure I wasn't crazy.
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