Proof using Triangle inequality

• Sep 29th 2009, 02:07 PM
eXist
Proof using Triangle inequality
I have the problem:

Prove:
$\displaystyle ||x| - |y|| \le |x - y|$

And I know I'm suppose to use the triangle inequality and the fact:
$\displaystyle x = x + y - y$ and $\displaystyle y = y + x - x$

I tried starting with:

$\displaystyle |x - y| =>$
$\displaystyle |(x - y + y) - (y - x + x)|$

But I don't know how to break this up using the triangle inequality:
$\displaystyle |x + y| \le |x| + |y|$
• Sep 29th 2009, 02:22 PM
Plato
$\displaystyle \left| x \right| = \left| {x - y + y} \right| \leqslant \left| {x - y} \right| + \left| { y} \right|\; \Rightarrow \;\left| x \right| - \left| y \right| \leqslant \left| {x - y} \right|$

Likewise: $\displaystyle \left| y \right| = \left| {y - x + x} \right| \leqslant \left| {y - x} \right| + \left| x \right|\; \Rightarrow \;\left| y \right| - \left| x \right| \leqslant \left| {y - x} \right|$

Recall that $\displaystyle |x-y|=|y-x|$

So putting the first two together we get.
$\displaystyle - \left| {x - y} \right| \leqslant \left| x \right| - \left| y \right| \leqslant \left| {x - y} \right|\; \Rightarrow \;\left| {\left| x \right| - \left| y \right|} \right| \leqslant \left| {x - y} \right|$.
• Sep 29th 2009, 03:03 PM
eXist
I'm so upset with myself. I honestly tried that but never thought of moving $\displaystyle |y|$ to the LHS. :/........

Thanks so much :D. You're my hero.