Consider:
The absolute value of |x − 1| − |x − 3| can be proven to equal
|(x − 1) − (x − 3)| {or |2|} at most.
Then write | |x − 1| − |x − 3| | ≤ |x − 1 − x + 3| which goes to
| |x − 1| − |x − 3| | ≤ 2.
The correct inequality (borne out by graphing calculator) would then be
-2 ≤ |x − 1| − |x − 3| ≤ 2 or | |x − 1| − |x − 3| | ≤ 2.
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The Triangle Inequality has it that |x − 1| expressed as
| (x − 3) + [(x − 1) − (x − 3)] | is at most |(x − 3)| + | [(x − 1) − (x − 3)] |.
Simplify this last to | (x − 1) | ≤ | (x − 3) | + | [2] |.
Take |x − 3| from both sides and rewrite the last inequality above as
|x − 1| − |x − 3| ≤ |2| which can go to | |x − 1| − |x − 3| | ≤ 2 as well.
