Define for non-negative a and prove two things:Originally Posted by hoeltgman
1. is increasing
Using also the triangle inequality you should be able to complete the proof.
We define a distance in like the following:
A distance is defined by the following properties:
1. d(x,y) >= 0
2. d(x,y) = d(y,x)
3. d(x,y) =< d(x,z) + d(z,y)
I managed to prove the two first (it was rather easy). But I'm stuck on the third. Does anybody know how to do? I tried various things from expanding the first part to reducing the second part with the triangular inegality or even the inequation of Cauchy-Schwartz. I ended up always in something that I couldn't prove anymore.
I would be thankful for any idea.
Originally Posted by hpe
I've got a few things that don't seem really clear to me. The first point is clear. I derivate and find out that it's always positive....
I do have problems to see how come from your second point to my original problem. If I say a = |x-y| then b could be |x-z| or |y-z|. Something is missing here I think. I tried your way and ended up with a + b + 1 -ab < 0. which doesn't really help since even if I transform back, I find can find the third distance, but I still have a term which keeps me from getting any good conclusion.
a = |x-y|, b =|y-z|, c = |x-z|. You want to proveOriginally Posted by hoeltgman
and know from the ordinary triangle inequality that .
because f is increasing
because of the second property. You are done. As to the proof of the second property...
Hope this helps