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.

you know that $|a+b|\leq |a|+|b|$, now take $a=x-y$ and $b=y$. — Yanko, Feb 19 '18 at 17:19
While not a short proof, this can be approached by considering several cases that allow you to remove the absolute values. — Dunham, Feb 19 '18 at 17:37
The standard way is yanko's comment and Maths survivor's answer. But I prefer cases: $a-b= \pm|a|\pm|b|$ and $|a-b| = \max(|a|,|b|)\pm\min(|a|,|b|) \ge \max(|a|,|b|) - \min(|a|,|b|) = ||a|-|b||$ — fleablood, Feb 19 '18 at 18:05
Thank You all. I get the answer and the right way to solve such problems too. — Ashish Gupta, Feb 23 '18 at 13:55

score 1 · Answer 1 · answered Feb 19 '18 at 17:33

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:

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

score 1 · Answer 2 · answered Feb 19 '18 at 17:40

1

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

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

Exchange roles of $a$ and $b$:

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

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

$\rightarrow:$

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

answered Feb 19 '18 at 17:40

Peter Szilas

20,344
2
17
28

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

2 Answers2