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## Proof: (related to "Reverse Triangle Inequalities")

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

### Proof of $(2)$

• 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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