# Homework Help: Using the generalized triangle inequality

1. Nov 3, 2012

### scharl4

1. The problem statement, all variables and given/known data
Using the generalized triangle inequality, prove |d(x,y) - d(z,w)| ≤ d(x,z) + d(y,w)

2. Relevant equations
d(x,y) is a metric
triangle inequality: d(x,y) ≤ d(x,z) + d(z,y)

3. The attempt at a solution
I know that this needs to be proved with cases: a) d(x,y) - d(z,w) ≥ 0, and b) d(x,y) - d(z,w) < 0. I know that for case a), we have |d(x,y) - d(z,w)| = d(x,y) - d(z,w). The part I am stuck at is that by applying the generalized triangle inequality to this expression, we are supposed to get d(x,y) - d(z,w) ≤ d(x,z) + d(w,y). I just do not understand why. Can anyone help please?

2. Nov 3, 2012

### micromass

Why would this prove anything?? Why do you even think (a) and (b) are true?

3. Nov 3, 2012

### scharl4

OK, I used the two cases a) d(x,y) - d(z,w) ≥ 0 and b) d(x,y) - d(z,w) < 0 because of the absolute value in the problem. So for case a) i need to prove d(x,y) - d(z,w) ≤ d(x,z) + d(y,w) and for case b) i need to prove d(z,w) - d(x,y) ≤ d(x,z) + d(y,w)

4. Nov 3, 2012

### micromass

OK, that makes sense.

So you need to prove d(x,y)-d(z,w) ≤ d(x,z) + d(y,w). What happens if you add d(z,w) to both sides?

5. Nov 3, 2012

### scharl4

Ok, so then adding d(z,w) to both sides gives d(x,y) ≤ d(x,z) + d(y,w) + d(z,w), so d(x,z) + d(z,w) + d(y,w) ≥ d(x,w) + d(w,y) by the triangle inequality, and d(x,w) + d(w,y) ≥ d(x,y) by the triangle inequality, so d(x,y) ≤ d(x,z) + d(y,w) + d(z,w). Thanks so much, now I understand where this is coming from!

6. Nov 3, 2012

### micromass

Note: if you actually write up the proof then you have to reverse it. You can't write: from $d(x,y)-d(z,w) \leq d(x,z) + d(y,w)$ follows $d(x,y)\leq d(x,z)+d(y,w)+d(w,z)$. This is true, but it isn't helpful since you started with what you wanted to proof.

What you have to do is write: from $d(x,y)\leq d(x,z)+d(y,w)+d(w,z)$ follows $d(x,y)-d(z,w) \leq d(x,z) + d(y,w)$.