# Triangle Inequality

• Sep 20th 2011, 03:14 PM
ekerik
Triangle Inequality
Hi, sorry if this is the wrong area I wasn't sure where to put it.

I did a question:

Determine whether the inequality is an identity. [Hint: Use the Triangle Inequality ( |a+b| <= |a| + |b|) with a = xy and b = y.]

|xy| ≥ |x| − |y|

I decided it was an identity, but not by using the hint (more like that it seemed logical but I can't write up a proof for why I think so - and that's not really good enough for me despite the fact that in this case it did get a correct answer). Anyways, my question is why do they suggest a = x - y and b = y ? I can see some relation in that the inequality has x - y in it but other than that I don't see why these were suggested, they seem arbitrary to me but I figure there must be a reason.

I'm not very experienced with math and the hint is the first time I've even heard of the Triangle Inequality, in fact I'm not even sure why the Triangle Inequality is being used to solve this, other than the clear similarity between the original Triangle Inequality equation and my current equation. So that's probably most of the problem...

Thanks in advance for any help with this!
• Sep 20th 2011, 03:32 PM
Plato
Re: Triangle Inequality
Quote:

Originally Posted by ekerik
Determine whether the inequality is an identity. [Hint: Use the Triangle Inequality ( |a+b| <= |a| + |b|) with a = xy and b = y.]
|xy| ≥ |x| − |y|

Here is the basic idea of this proof.
\$\displaystyle -a\le b\le a\$ implies that \$\displaystyle |b|\le |a|\$.

Now we have
\$\displaystyle |a|\le |a-b|+|-b|=|a-b|+|b|\$ so \$\displaystyle |a|-|b|\le |a-b|\$

Likewise we know \$\displaystyle -(|a|-|b|)=|b|-|a|\le |b-a|=|a-b|\$

So apply that basic idea.
• Sep 21st 2011, 06:30 AM
HallsofIvy
Re: Triangle Inequality
You don't really need to note that \$\displaystyle -a\le b\le a\$. From \$\displaystyle |x+ y|\le |x|+ |y|\$, with x= a- b, y= b, we have immediately \$\displaystyle |a- b+ b|= |a|\le |a- b|+ |b|\$. Now subtract |b| from both sides.