Question

Proof of the triangle inequality on Euclidean n-space.

Answer 1

google, get maybe a half dozen proofs, pick the one you like and understand the best

Answer 2

Not sure why I'm getting so mixed up on this. I'm trying to prove:\[
|z-x|\le|z-y|+|y-x|
\]with \(|x|\) defined as \[\sqrt{\sum_{i=1}^n x_i^2}\]@satellite73 I've proved the triangle inequality before, that's not the problem. The problem is proving it in this certain context.

Answer 3

We can also use inner products here, \[\left<x,y\right>=\sum_{i=1}^n x_iy_i\]

Answer 4

Also, in general, "google to find the answer" is a terrible response on OpenStudy, satellite. You ought to know better than that.

Answer 5

My main question here is, I'm not sure if I'm missing an easy step or if the proof is actually just really really messy. I'm ending up fully expanding the term \(|z-y|\cdot|y-x|\), which leads to a huge mess. My previous proofs of this inequality (in much less abstract contexts) were never this messy.

Answer 6

|z-x| = |z-y+y-x| = |(z-y)+(y-z)|

Answer 7

Oooh that might help. Let me play with that for a moment.

Answer 8

Once you have those two parenthesis, you can just use normal triangle inequality logic.

Answer 9

So that basically reduces it to \(|x+y| \le |x|+|y|\) which I already proved.

Answer 10

Awesome, that's the trick I was looking for. Thanks Smooth.

Answer 11

Oh, typo.
|z-y+y-x| = |(z-y)+(y-x)|

Answer 12

Yup =) My pleasure.