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Analysis problem

  1. Sep 1, 2007 #1
    My problem is the next:

    Show that [tex]\forall x,y\geq 0[/tex] and [tex]\frac{1}{p}+\frac{1}{q}=1[/tex] we have that [tex]xy\leq\frac{x^{p}}{p}+\frac{y^{q}}{q}[/tex].

    I use the logarithm function because it is concave and grows monotonically and then


    the problem is I don't know to justify the next step (or is justified yet?)

    [tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}\leq \ln\left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right)[/tex].

    That's all, I wait for your help.
  2. jcsd
  3. Sep 1, 2007 #2


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    Staff Emeritus
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    Gold Member

    Derivatives are often good at proving inequalities. e.g. you can prove
    f(x) < g(x)
    for all positive x as follows:
    . f(0) < g(0)
    . f'(x) < g'(x) for all positive x

    (I don't know if they will be useful here, but it's something I'd try)
  4. Sep 3, 2007 #3
    This may be brute force but it looks like it works:
    Use Arithmetic Mean>= Geometric Mean.
    1/p+1/q=1 implies p+q=p.q
    Also x^p/p+y^q/q=(q.x^p+p.y^q)/(p.q)
    Now AM>=GM implies
    (q instances of x^p plus p instances of y^q)/(p+q)>={x^(p.q).y^(q.p)}^(1/(p+q))
    LHS is just (p.q).{x^p/p+y^q/q}/(p+q)=x^p/p+y^q/q
    RHS is x.y

    Hope this helps
    Last edited by a moderator: Sep 3, 2007
  5. Sep 3, 2007 #4

    Gib Z

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    Homework Helper

    Hurkyl, Derivatives only help easily in inequalities of 1 variable :( Even if he expressed y as a function of x, the condition on p and q pose difficulty.
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