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Putnam problem with inequalities

  • Thread starter ehrenfest
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  • #1
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Homework Statement


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?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #4
D H
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You want to leave expressions involving {ri} on both sides of the inequality.
 
  • #5
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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?
 
  • #6
D H
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You don't need Cauchy-Schwartz. You are on the right track. Set each xi to ri.
 
  • #7
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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 :(
 
  • #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.
 
  • #9
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When r is a null vector its length is certainly less than or equal to 1.
 
  • #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.
 
  • #11
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I agree. Thanks.
 

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