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Putnam problem with inequalities

  1. Nov 9, 2007 #1
    1. The problem statement, all variables and given/known data
    can someone help me with only if part of Problem 2?

    http://www.math.cornell.edu/~putnam/ineqs.pdf

    I can prove that quantity is less than or equal to n, but I cannot get it leq to 1?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2007 #2

    D H

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    Assume the first expression is true for all [itex]x_1, x_2, \cdots, x_n[/itex]. This means you can hand-pick a particular [itex]x_1, x_2, \cdots, x_n[/itex] that makes the desired relation fall out.
     
  4. Nov 9, 2007 #3
    Yes. That's how I got that it was less than n. I just picked basis vectors. But I don't know which special vectors pick to get it less than 1.
     
  5. Nov 9, 2007 #4

    D H

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    You want to leave expressions involving {ri} on both sides of the inequality.
     
  6. Nov 9, 2007 #5
    So, should I let x equal r times a scalar factor or something? I need to use the Cauchy-Schwartz inequality, somehow, don't I?
     
  7. Nov 9, 2007 #6

    D H

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    You don't need Cauchy-Schwartz. You are on the right track. Set each xi to ri.
     
  8. Nov 9, 2007 #7
    So then we have that norm(r) => norm(r)^2 i.e. that 1 => norm(r). I cannot believe how long I spent on that :(
     
  9. Nov 9, 2007 #8

    D H

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    Almost. You get norm(r)^2 >= norm(r)^4, and thus 1 >= norm(r)^2.

    To be completely rigorous you also have to handle the special case where the division in the final step is not valid. This case is trivial, so to speak.
     
  10. Nov 9, 2007 #9
    When r is a null vector its length is certainly less than or equal to 1.
     
  11. Nov 9, 2007 #10

    D H

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    Its trivial to deal with the zero vector, but you deal with it you must as step from norm(r)^2 >= norm(4)^r to 1>= norm(r)^2 is not valid for r=0.
     
  12. Nov 9, 2007 #11
    I agree. Thanks.
     
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