By hypothesis, $x,y\in\mathbb R$ are arbitrary real numbers.

- Let $u:=x+y$ and $v:=-y$
- By the triangle inequality, we have $$|u+v|\le |u|+|v|$$
- By definition $$|x|\le|x+y|+|-y|=|x+y|+|y|.$$
- By rule 3 for calculations with inequalities we can add $-|y|$ on both sides to get $$|x+y|\ge |x|-|y|.$$

- From $(1)$ it follows $|x-y|=|x+(-y)|\ge |x|-|(-y)|=|x|-|y|.$
- For the same reason $|y-x|\ge |y|-|x|.$
- By definition of absolute value $|x-y|=|y-x|.$
- Thus, both inequalities hold simultaneously: $|x-y|\ge |x|-|y|$ and $|x-y|\ge |y|-|x|.$
- The greater of $|x|-|y|$ and $|y|-|x|$ equals $\left||x|-|y|\right|.$
- Therefore, $$|x-y|\ge \left||x|-|y|\right|.$$

q.e.d

| | | | created: 2020-02-06 20:20:20 | modified: 2020-02-06 20:20:37 | by: *bookofproofs* | references: [581]

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[581] **Forster Otto**: “Analysis 1, Differential- und Integralrechnung einer Veränderlichen”, Vieweg Studium, 1983