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 $d(x,y)-d(z,w) \leq d(x,z) + d(y,w)$.