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Need help with a complex inequality?

  1. Sep 26, 2004 #1
    Need help with a complex inequality??

    hey!!
    i been trying to do this inequality for a 2 hrs now and cant seem to prove it
    [tex]|\frac{1}{2}(a+b)|^p \leq \frac{1}{2}(|a|^p+|b|^p)[/tex] where a,b are complex numbers
    Can anyone suggest a way??
    thanks
     
    Last edited: Sep 26, 2004
  2. jcsd
  3. Sep 26, 2004 #2
    Try this:

    [tex]|\frac{1}{2}(a+b)|^{p}\leq(\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p}-pab[/tex]

    Yes, I know, the last term is only pab if p=2, but you will always be subtracting somthing at the end, no matter the value of p. I think this kinda works...
     
  4. Sep 26, 2004 #3
    so you saying that [tex]|\frac{1}{2}(a+b)|^{p}\leq(\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p}-pab \leq \frac{1}{2}(|a|^p+|b|^p)[/tex]
     
  5. Sep 26, 2004 #4
    for p = 1
    [tex]|\frac{1}{2}(a+b)|\leq(\frac{1}{2}|(a+b)|)-ab [/tex]
    isnt this false beacause you subtracting a ab on the RHS???
     
    Last edited: Sep 26, 2004
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