# triangle inequality

• Jul 17th 2013, 12:03 PM
jys89
triangle inequality
Hello,
I had a question regarding triangle inequalities.
I proved the inequality abs(x+y) <= abs(x) + abs(y).

Second part of question is asking to use the triangle inequality to prove "shortest path" version:
for all x,y,z in R, abs(x-y) <= abs(x-z) + abs(z-x).
I've got no idea on how to start this part.
Should there be a y in the right half of the inequality? Am I supposed to assume something about y S.T it doesn't need to be
in the right half?
Thanks all
• Jul 17th 2013, 12:17 PM
emakarov
Re: triangle inequality
I assume the inequality should be

abs(x - y) <= abs(x - z) + abs(z - y).

Note that the abs arguments in the right-hand side add up to the abs argument in the left-hand side: (x - z) + (z - y) = x - y. This is similar to the original triangle inequality. Does this help?
• Jul 17th 2013, 12:19 PM
Plato
Re: triangle inequality
Quote:

Originally Posted by jys89
I proved the inequality abs(x+y) <= abs(x) + abs(y).

Second part of question is asking to use the triangle inequality to prove "shortest path" version:
for all x,y,z in R, abs(x-y) <= abs(x-z) + abs(z-x).

As written it is false. Counterexample: \$\displaystyle x=1,~y=5,~\&~z=2\$.

You can show that \$\displaystyle |x-y|\le|y-z|+|z-y|\$ simply by noting that \$\displaystyle (x-y)=(x-z)+(z-y),\$
• Jul 17th 2013, 12:26 PM
jys89
Re: triangle inequality
Thanks to both of you guys. I knew something about the problem was wrong.