Prove that $||a|-|b|| \leq |a-b|$?

Question

How do we prove that $||a|-|b|| \leq |a-b|$ ? I do know that $|a-b|<|a|+|b|$ from the triangle inequalities. As a matter of fact it does remind me of something, I studied in Complex numbers class, Something like this.

However, I can't relate both. I did also tried to expand the inequality, with no success, just too many cases for me to handle. Hopefully I can get an answer here. Thanks.

Answer 1

You know that $|x+y|\leq |x|+|y|$ holds for every real number , then taking $x=a-b$ and $y=b$ we get:

$|a-b|\geq|a|-|b|$ ... (1)

if we exchange $a$ and $b$ we'll have:

$|a-b|\geq -(|a|-|b|)$, or

$-|a-b|\leq|a|-|b|$ ... (2)

Now from (1) and (2) based on absolute value properties we get:

$||a|-|b||\leq|a-b|$

Answer 2

$|a| =|a-b+b| \le |a-b| +|b|$, or

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

Exchange roles of $a$ and $b$:

$|b|=| b-a+a| \le |b-a| +|a|$, or

$|b|-|a| \le |b-a|= |a-b|$.

$\rightarrow:$

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