Looking for a simple generalization regarding absolute values

[1MileCrash]

Hi all,

I was working on a proof that essentially worked because:

|x-y|+|y-z| >= |x-y+y-z|

I knew this was true because, but I'm looking for a generalization in a way that I could write in a proof.

Can you say that when comparing two expressions of addition/subtraction that are identical except for absolute values, if one has more absolute value symbols, it is always greater (or equal)? But then that would also require me to say something like "it has to be a sum/can't be a "negative absolute value (subtraction).." It's hard to explain.

How would you justify the step I did?

 

[Office_Shredder]

Do you mean like
\left| \sum_{i=1}^{n} a_i \right| \leq \sum_{i=1}^{n} |a_i|

Your step is literally the triangle inequality. Let a_1 = x-y and a_2 = y-z then you just wrote that
|a_1 + a_2| \leq |a_1| + |a_2|

 

[1MileCrash]

Do you mean like
\left| \sum_{i=1}^{n} a_i \right| \leq \sum_{i=1}^{n} |a_i|

Your step is literally the triangle inequality. Let a_1 = x-y and a_2 = y-z then you just wrote that
|a_1 + a_2| \leq |a_1| + |a_2|

Yes, it is the triangle inequality, but I was trying to prove that something was a metric, which required me to prove that the triangle inequality held true for that function. Am I allowed to say "the triangle inequality holds in this metric due to the triangle inequality" in this case since it reduces to the standard triangle inequality? It sounds... weird.

 

[HallsofIvy]

Yes, you can say that if you also show that it reduces to the "standard" triangle inequality (by which I assume you mean with real numbers).

 

[1MileCrash]

Yes, real numbers. These numbers are components of a member of R^n, X, Y, and Z.

I think I did show it, basically those expressions are both in summations as part of the metric. And due the standard triangle inequality, the triangle inequality for the entire metric (with summation) holds, because for any ith term in the summation, one is surely greater due to the regular triangle inequality. So the sum is, too.

Or something like that.

Thanks all