Inverse triangle equality

Question

Possible Duplicate:
Why exactly can you take the absolute value of one side of this inequality and assume it is still true?

Why is $||a|-|b|| \ge |a|-|b|$, tried a lot (like comparing to quadratics), but couldn't find the answer. Can anyone help me in the right direction?

Regards, Kevin

Answer 1

Apply the common triangle inequality $$ \|x\| + \|y\| \geq \|x+y\| $$ to the variables $x=(|a|-|b|)$ and $y=|b|$

Answer 2

HINT: Writing $x = |a| - |b|$, the expression reduces to $|x| \geq x$...

Answer 3

You want to prove $|a| - |b| \le ||a| - |b||$.

It is very simple! No triangle equality complications! Just simply split it into two cases:

1. $|a| - |b| \le 0$, then the theorem is obvious because clearly $0 \le ||a| - |b||$.
2. $|a| - |b| > 0$, then $||a| - |b|| = |a| - |b|$.

QED.

Answer 4

Thanks. Stupid me! It was indeed $||a|-|b||\le|a-b|$

I did it this way, is it OK?

$||a|-|b||^2=(|a|-|b|)^2=|a|^2+|b|^2-2|a||b|$

$=a^2+b^2-2|ab|$

$\le a^2+b^2-2ab$

$=(a-b)^2=|a-b|^2$

So:

$||a|-|b||^2 \le |a-b|^2$

And therefor:

$||a|-|b|| \le |a-b|$