How to prove $||A| - |B|| \le |A -B|$

Question

Suppose A and B are real numbers, prove following inequality: $$ ||A| - |B|| \leq |A - B| $$

How to prove this inequality？ Thanks.

Check out Wikipedia for 'reverse triangle inequality' there is a proof given. — bonanza, May 24 '14 at 17:44
Why not consider the cases where both are positive, one positive and one negative, or both are negative? Then recall that $\vert A \vert = A$ when $A\geq0$ and $-A$ when $A<0$. — Philip Hoskins, May 24 '14 at 17:46
http://math.stackexchange.com/questions/774197/intuition-or-figure-for-reverse-triangle-inequality-mathbfa-mathbfb/774233#774233 — S L, May 24 '14 at 17:51
It turns out this has already been solved in math.se, in the following link:http://math.stackexchange.com/questions/127372/reverse-triangle-inequality-proof — wangshuaijie, May 24 '14 at 17:56

score 7 · Accepted Answer · answered May 24 '14 at 17:52

7

\begin{align} |P+Q| & \le |P|+|Q| \\[8pt] \text{Therefore }|(A-B)+B| & \le |A-B| + |B|. \\[8pt] |A| & \le |A-B|+|B| \\[8pt] |A| - |B| & \le | A - B| \end{align}

By the same method, show that $|B|-|A|\le|A-B|$ and then you've got it.

answered May 24 '14 at 17:52

Michael Hardy

1

score 2 · Answer 2 · answered May 24 '14 at 17:57

Note that $|x-y|$ is the distance between the points $x$ and $y$ on the real number line. We have to prove that $|A|$ and $|B|$ are at least as close as $A$ and $B$. If $A$ and $B$ are of the same sign, $|A|$ and $|B|$ are as close as $A$ and $B$. If they are of different signs, $|A|$ and $|B|$ are closer than $A$ and $B$ since the distance between $|A|$ and $|B|$ now equals the distance between $A$ and $-B$ (assuming $|A|>|B|$) where $-B$ lies between $A$ and $B$ and therefore is closer to $A$ than $B$ is.

score 1 · Answer 3 · answered May 24 '14 at 18:01

This is called the inequality of the uniform continuty of $|\cdot|$ in a more general normed linear space than $\mathbb{R}$ any norm satisfies that condition. For alternative proof, squaring both sides to obtain and equivalent inequality $$-2|A||B|\leq 2AB$$ which is trivially true.

But i like more the Hardy's answer :)

How to prove $||A| - |B|| \le |A -B|$

3 Answers3