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Help proving triangle inequality for metric spaces

  1. Sep 12, 2015 #1
    So, i need to proof the triangle inequality ( d(x,y)<=d(x,z)+d(z,y) ) for the distance below
    5XzC654.jpg
    But i'm stuck at
    S6sjolB.jpg

    In those fractions i need Xk-Zk and Zk-Yk in the denominators, not Xk-Yk and Xk-Yk. Thanks in advance
     
  2. jcsd
  3. Sep 12, 2015 #2

    mfb

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    a/(1+a) = 1 - 1/(1+a)
    Then you just have to transform the denominator and not both parts. Not sure if that is the best approach, but it is the one I would try first.
     
  4. Sep 13, 2015 #3
    I used somthing similar instead Minkowski, i did (a+b)/(1+a+b) <= a/(1+a) + b/(1+b)
    0Ko8ooz.jpg f
     
  5. Sep 13, 2015 #4

    mfb

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    I don't think that is a valid transformation with the denominator. You change 1+|g+f| to 1+|g|+|f| which might increase the fraction.
     
    Last edited: Sep 13, 2015
  6. Sep 13, 2015 #5
    But in that case g and f are > 0, so in no way will reduce the fraction. |g|+|f|>|g+f| only if g or f is <0
     
  7. Sep 13, 2015 #6

    mfb

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    You don't know that about f and g (which are the raw differences, e.g. xk-zk).

    "reduce" in the last post should be "increase", of course.
     
  8. Sep 13, 2015 #7
    (Thank you for all your help s far) What if i told you that |a-b+b-c| = |a-b| + |b-c|? I'm having a little trouble prooving that but i'm pretty confident.
     
  9. Sep 13, 2015 #8

    mfb

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    Not in general.

    a=2, b=5, c=1
    1 = 3 + 4?
     
  10. Sep 15, 2015 #9
    You should use the fact that the function x/(1+x)=1-1/(1+x) is increasing.
     
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