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Larsen problem

  1. Dec 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Is what I wrote on the left hand margin a counterexample to 1.8.5 part a) ?

    EDIT: I meant part b)

    2. Relevant equations



    3. The attempt at a solution
     

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    Last edited: Dec 24, 2007
  2. jcsd
  3. Dec 24, 2007 #2

    Shooting Star

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    Yes. (For b).
     
  4. Dec 24, 2007 #3
    And changing a+b=1 to a+b=2 makes the inequality correct, right?

    Also, in part b) of 1.8.7, are they asking me to show that 1/(a+c) < 1/(b+c) + 1/(a+b) ?
     
  5. Dec 24, 2007 #4

    Gokul43201

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    At a+b=2, you will not see the equality. There's a lower value of a+b that will realize this.
     
  6. Dec 24, 2007 #5

    Gokul43201

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    I think not (it's poorly written).

    I think they want you to show : 1/(b+c) < 1/(a+c) + 1/(a+b)
     
  7. Dec 24, 2007 #6
    How is that different? The problem is symmetric in a,b,and c; they are just the lengths of the three sides of a triangle.

    What should the problem be then?
     
    Last edited: Dec 24, 2007
  8. Dec 24, 2007 #7

    Gokul43201

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    Doh! Right - I can't remember what I was thinking.

    Several possibilities. For one, a+b=\sqrt{2}, everything else unchanged.
     
  9. Dec 24, 2007 #8
    For 1.8.5, with Gokul's correction, I still don't see how to prove it. I find that the inequality is equivalent to:

    [tex] 2 \leq a^2+b^2+ab(y/x +x/y) [/tex]

    y/x+x/y is greater than or equal to 2, but I do not see how that helps.

    For 1.8.6, I found that the inequality is equivalent to:

    [tex] b^2 < a^2 + b(a+c)+ac+c^2[/tex]

    which I am not sure how to prove.
     
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