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AMGM(n) help

  1. Nov 17, 2007 #1
    1. The problem statement, all variables and given/known data
    http://math.stanford.edu/~vakil/putnam07/07putnam1.pdf

    Can someone help me with problem 8 part b?

    I plugged that in to AMGM(n) and the LHS easily transforms into the LHS of AMGM(n-1) but I am not sure how the RHS transforms to the RHS of AMGM(n-1). Specifically, what happens to the exponent?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 17, 2007 #2
    This is the famous Cauchy proof of Am-Gm. Post what you dd so far.
     
  4. Nov 17, 2007 #3
    Again, I plugged in the hint to AMGM(n) and the LHS easily transforms into the LHS of AMGM(n-1).

    So I have the LHS of AMGN(n-1) is greater than or equal to [tex](\frac{a_1...a_{n-1} (a_1+...+a{n-1})}{(n-1)})^{1/n}[/tex]

    and I do not know where to go from there.
     
  5. Nov 18, 2007 #4

    morphism

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    Work with
    [tex]\left( \frac{\sum a_i}{n} \right)^n \geq \prod a_i[/tex]
    instead.

    Keep an eye on the possibility of canceling something convenient off both sides after you apply the hint.
     
  6. Nov 18, 2007 #5
    After I apply the hint, it looks like this:

    [tex]\left( \frac{\sum_{i=1}^{n-1} a_i}{n-1} \right)^n \geq 1/(n-1)\sum_{i=1}^{n-1} a_i \prod_{i=1}^{n-1} a_i[/tex]

    which is the same as

    [tex]\left( \frac{\sum_{i=1}^{n-1} a_i}{n-1} \right)^{n-1} \geq \prod_{i=1}^{n-1} a_i[/tex]

    because everything is nonnegative.

    On to part c).
     
    Last edited: Nov 18, 2007
  7. Nov 18, 2007 #6
    And of course I got stuck on part c) as well. I tried doing something similar like letting a_i = a_i + a_{n+i} or a_i = a_i *a_{i+n} but that only works for one side at a time. I tried to do the case k=2 "by hand" but there is the (a_1 +a_2+a_3+a_4)^4 part that is prohibitive. And I tried other stuff that was just as bad. So I am out of ideas.
     
  8. Nov 18, 2007 #7

    morphism

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    Here, let's assume k=2 holds and show k=4.

    [tex](a_1 a_2 b_1 b_2)^{1/4} = \left( (a_1 a_2)^{1/2} (b_1 b_2)^{1/2} \right)^{1/2} \leq \frac{(a_1 a_2)^{1/2} + (b_1 b_2)^{1/2}}{2} [/tex]

    Can you take it from here?
     
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