# Looking for a simple generalization regarding absolute values

1. Jun 20, 2013

### 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?

2. Jun 20, 2013

### Office_Shredder

Staff Emeritus
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
$$|a_1 + a_2| \leq |a_1| + |a_2|$$

3. Jun 20, 2013

### 1MileCrash

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.

4. Jun 21, 2013

### 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).

5. Jun 21, 2013

### 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

Last edited: Jun 21, 2013