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Help me prove this inequality

  1. Nov 9, 2007 #1

    quasar987

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    [SOLVED] Help me prove this inequality

    1. The problem statement, all variables and given/known data
    The inequality in question is

    [tex]|x+y|^p \leq 2^p(|x|^p+|y|^p)[/tex]

    for any positive integer p and real numbers x,y.

    3. The attempt at a solution
    For p=1, it is weaker than the triangle inequality.

    Suppose it is true for p, and let's try to show this implies it's true for p+1.

    [tex]|x+y|^{p+1}=|x+y||x+y|^p\leq |x+y|2^p(|x|^p+|y|^p)[/tex]

    And basically, here I've tried using the triangle inequality on |x+y| but the most "reduced form" I got is I arrived at the conclusion that the inquality was true iff

    [tex]|x||y|(|x|^p+|y|^p)\leq |x|^{p+1}+|y|^{p+1}[/tex]
     
    Last edited: Nov 10, 2007
  2. jcsd
  3. Nov 9, 2007 #2
    Let [tex]a,b\geq 0[/tex] we want to show [tex](a+b)^n \leq 2^n(a^n+b^n)[/tex]. First accept induction. Then [tex](a+b)^{n+1} = (a+b)^n(a+b)\leq 2^n(a^n+b^n)(a+b) = 2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)[/tex]. But [tex]2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)\leq 2^{n+1}(a^{n+1}+b^{n+1})[/tex] iff [tex]a^{n+1}+b^{n+1}\leq 2(a^{n+1}+b^{n+1}+a^nb+ab^n)[/tex] iff [tex]a^{n+1}+b^{n+1}\geq ab^n+a^nb[/tex] iff [tex](a-b)(a^n-b^n)\geq 0[/tex] but that is true because [tex](a-b)(a^n-b^n) = (a-b)^2(a^{n-1}b+...+ab^{n-1})\geq 0[/tex].
     
  4. Nov 9, 2007 #3

    quasar987

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    Ok, this is just a typo probably because you fall back on your feet a few lines later with

    [tex]a^{n+1}+b^{n+1}\geq ab^n+a^nb[/tex]

    Good work, thanks Kummer.
     
    Last edited: Nov 9, 2007
  5. Nov 10, 2007 #4

    quasar987

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    How do I add [SOLVED] to the title?
     
  6. Nov 10, 2007 #5
    Click on "thread tools".
     
  7. Nov 10, 2007 #6

    Kurdt

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