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I wonder if $\lvert a-b\rvert\le\lvert a\rvert+\lvert b\rvert$ is always true.

I think it is true, but I don't see how to prove this mathematically.

Thanks.

YORRAY
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4 Answers4

7

Assume $|a+b|\leq |a|+|b|$. Then $|a-b|=|a+(-b)|\leq |a|+|(-b)|=|a|+|b|$.

3

$$|a-b|=|a+(-b)|\leq |a|+|-b|=|a|+|b|$$

Surb
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2

By the triangle inequality $|a+b|\le |a|+|b|$, so also, $$|a-b|=|a+(-b)|\le |a|+|-b|=|a|+|b|$$

J.R.
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1

If you want to prove the triangle inequality, consider proving $$|a+b|\leq |a|+|b|$$

when $\{a\leq 0, b\geq 0$}, $\{a,b \geq 0\}$, and then reason why that would also cover the cases $\{a\geq 0, b\leq 0\}$ and $\{a,b\leq 0\}$.