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Proof

(related to Theorem: Reverse Triangle Inequalities)

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

Proof of $(1)$

• 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 $$|x+y|\ge |x|-|y|.$$

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, $$|x-y|\ge \left||x|-|y|\right|.$$
  ∎

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983