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Proof question 2,1

  1. Oct 26, 2009 #1


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    it is given that
    [tex]1\leq p< +\infty\\[/tex]
    [tex]\alpha ,\beta >0 \\[/tex]
    [tex]a,b\geq 0\\[/tex]

    prove that
    [tex](\alpha a+\beta b )^p\leq (\alpha +\beta )^p\left ( \frac{\alpha }{\alpha +\beta }a^p+\frac{\beta }{\alpha +\beta }b^p \right )[/tex]

    hint: prove first that [tex]f(t)=t^p[/tex] is a convex
    on this region [tex][0,+\infty)[/tex]
    reminder: function f(t) is called convex on some region if for every b,a
    and on
    [tex]0\leq \lambda \leq 1\\[/tex]
    we have
    [tex]f(\lambda a +(1-\lambda)b)\leq\lambda f(a)+(1-\lambda)f(b)[/tex]

    my thoughts:
    i know from calc1 that a function is convex if its second derivative is negative or something (i am not sure)

    i dont know
    how to prove that [tex]f(t)=t^p[/tex] is negative
    its pure parametric thing

  2. jcsd
  3. Oct 26, 2009 #2


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    Science Advisor
    Homework Helper

    You've got it backwards. f(t) is convex if the second derivative is positive, not negative. So you don't want to show f(t) is negative, you want to show f''(t) is positive.
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