$||a|-|b|| \leq |a-b|$ - vector equivalent

Question

Let $\left|\cdot\right|$ be absolute value of a real number. Let $\langle\cdot\rangle$ be two-dimensional, Euclidean vector's norm.

Let a$\, =(x_1,y_1)$ and let b $=(x_2,y_2)$ .

I have the result that $||r_1|-|r_2||\leq |r_1-r_2|$ for $r_1,r_2\in\mathbb R$.

How do I prove that $\big|\big<$a$\big>-\big<$b$\big>\big|\leq\big<$a-b$\big>$?

This is the reverse triangle inequality ... a proof is given here https://math.stackexchange.com/questions/127372/reverse-triangle-inequality-proof — K Split X, Apr 18 '17 at 14:04

Robin Vogel · Accepted Answer · 2017-04-18T14:38:17.817

2

So, I note the euclidean norm $\|\cdot \|$, and the scalar product $\langle \cdot , \cdot\rangle$ $$ \|a-b\|^2 = \langle a-b, a-b \rangle = \|a\|^2 + \|b\|^2 - 2\langle a, b \rangle. $$ Or, using the Cauchy-Schwartz inequality $$ \langle a, b \rangle \le | \langle a, b \rangle | \le \| a\| \|b\| $$ So $$ \|a-b\|^2 \ge (\|a\| - \|b\|)^2 $$ And hence $$ \|a - b\| \ge | \|a\| - \|b\| |. $$

edited Apr 18 '17 at 14:38

answered Apr 18 '17 at 14:06

Robin Vogel

166

Just a formatting hint: Using | instead of || makes nicer norm bars (closer together). This becomes important when mixing with absolut value signs. Compare $||a|-|b||$ with $|||a||-||b|||$. Apart from that: you probably have mistyped your last inequality. – Florian Apr 18 '17 at 14:11
Thanks, didn't know that, edited ! – Robin Vogel Apr 18 '17 at 14:14
Much easier to read this way. ;-) – Florian Apr 18 '17 at 14:15
You have a typo in your last equation. Change the left side accordingly. – JJR Apr 18 '17 at 14:18
Yes, thank you, mistake changing the $|$. – Robin Vogel Apr 18 '17 at 14:39

$||a|-|b|| \leq |a-b|$ - vector equivalent

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