Proof of the triangle inequality on Euclidean n-space.

Question

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Yup =) My pleasure.