prove $|A+B|\geq|A|-|B|$

Question

Here is what I tried: $-|A| \leq A \leq |A|$ and $-|B|\leq B \leq |B|$ and adding the two inequality to get $|A+B| \leq |A|+|B|$.

How about $|A| = |(A+B) - B| \le |A+B| + |-B| = |A+B| + |B|$ — Umberto P., Feb 05 '20 at 14:34
Do you want to prove $|A+B|\geq|A|-|B|$ (Title) or $|A+B|\leq|A|+|B|$ (Text) ? — Andreas, Feb 05 '20 at 14:36
Probably some clarification could be useful. What are $A$ and $B$ (real numbers, complex numbers, sets)? What does the symbol $|\cdot|$ denote? (If they are numbers, most likely it is absolute value. If they are sets, most likely it is the number of elements.) — Martin Sleziak, Feb 05 '20 at 15:16
If the question is about numbers, the this is very close to reverse triangle inequality. See this post: Reverse Triangle Inequality Proof (and [other questions linked there). — Martin Sleziak, Feb 05 '20 at 15:17

score 2 · Accepted Answer · answered Feb 05 '20 at 14:37

2

$$|A+B|+|B|=|A+B|+|-B|\ge|A+B+(-B)|=|A|$$

answered Feb 05 '20 at 14:37

timon92

11,329

score 1 · Answer 2 · answered Feb 05 '20 at 14:42

If $A$ and $B$ are positve then the absolute value of their sum is greater then the difference $|A| - |B|$. Now, suppose $A$ positive and $B$ negative with $|A|>|B|$: in thos case we have $|A+B|=|B|$. From here it's very simple to show the inequality holds for any $A$ and $B$.

prove $|A+B|\geq|A|-|B|$

2 Answers2