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Is the following inequality(that looks like the triangle inequality) valid:

$|a - b| \leq |a| - |b|$

Why?

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    it would be also nice to see an intuitive explanation/answer to this. It feels it should be obvious, no?2017-11-02

3 Answers 3

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It's sometimes called the reverse triangle inequality. The proper form is $\left| a - b \right| \ge \big||a| - |b|\big|$ For the proof, consider $|a| = |a - b + b| \le |a - b| + |b|$ $|b| = |b - a + a| \le |a - b| + |a|$ so that we have $-|a-b|\le|a|-|b| \le |a - b|$

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    @CharlieParker It depends on what you mean by intuitive. The statement is a formalization of the fact that "the difference of two sides of a triangle is always less than (or equal to) the third side". I don't know if you find that fact intuitive or not, but it is just a restatement of the fact that "the sum of two sides of a triangle is always greater (or equal to) the third side", which is the triangle inequality itself. This is of course reflected in the fact that the reverse triangle inequality is a direct consequence of the triangle inequality.2017-11-02
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No. For example, $|(-2)-3|=5>|-2|-|3|=-1.$

I think you're thinking of $||a|-|b||\le |a- b|.$

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The length of any side of a triangle is greater than the absolute difference of the lengths of the other two sides:

$||a|-|b||\leq |a-b|$

Here is a proof:

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

and,

(1) $|a-b|\geq |a|-|b|$

Interchanging $a$ and $b$, we get also

(2) $|a-b|\geq |b|-|a|$

Combining (1) and (2) we get our desired result.