Help proving triangle inequality for metric spaces

[lucasLima]

So, i need to proof the triangle inequality ( d(x,y)<=d(x,z)+d(z,y) ) for the distance below

But I'm stuck at

In those fractions i need Xk-Zk and Zk-Yk in the denominators, not Xk-Yk and Xk-Yk. Thanks in advance

 

[mfb]

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.

 

[lucasLima]

I used somthing similar instead Minkowski, i did (a+b)/(1+a+b) <= a/(1+a) + b/(1+b)

f

 

[mfb]

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.

 

[lucasLima]

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

 

[mfb]

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

"reduce" in the last post should be "increase", of course.

 

[lucasLima]

(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.

 

[mfb]

Not in general.

a=2, b=5, c=1
1 = 3 + 4?

 

[Hawkeye18]

You should use the fact that the function x/(1+x)=1-1/(1+x) is increasing.