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Simple inequality?

  1. Nov 23, 2011 #1
    Is there a way to do this without differentiation?

    [itex]\left(a+b\right)^{p}[/itex] [itex]\leq a^{p}+b^{p}[/itex]


    0<p<1 and a,b[itex]\geq[/itex] 0

    pulling the a out of the the first part and dividing by it to get

    [itex]\left(1+\frac{b}{a}\right)^{p}[/itex][itex]\leq 1+\frac{b}{a}^{p}[/itex]

    This seems like the way to go but am stuck. Any suggestions? Thanks.
     
  2. jcsd
  3. Nov 23, 2011 #2
    Use the binomial theorem?
     
  4. Nov 23, 2011 #3
    p is not an integer though. Not sure binomial thm would work.

    0 < p < 1
     
  5. Nov 23, 2011 #4
    It works.
     
  6. Nov 23, 2011 #5
    Did not know that, will do some research.
     
  7. Nov 23, 2011 #6
    It is Newton's generalisation that works. It is there on wikipedia.
     
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