Help proving triangle inequality for metric spaces

  • #1
17
0
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
 

Answers and Replies

  • #2
34,783
10,943
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.
 
  • #3
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I used somthing similar instead Minkowski, i did (a+b)/(1+a+b) <= a/(1+a) + b/(1+b)
0Ko8ooz.jpg
f
 
  • #4
34,783
10,943
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:
  • #5
17
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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 reduce the fraction.
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
 
  • #6
34,783
10,943
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.
 
  • #7
17
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(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.
 
  • #8
34,783
10,943
Not in general.

a=2, b=5, c=1
1 = 3 + 4?
 
  • #9
177
61
You should use the fact that the function x/(1+x)=1-1/(1+x) is increasing.
 

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