# Triangle inequality for subtraction?

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

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

Why?

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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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Let $s_n$ be a sequence. Is this valid then: $|s_n - s| < 1 \iff ||s_n| - |s|| < 1 \iff |s_n| < |s| + 1$? – CodeKingPlusPlus Oct 15 '12 at 4:08
That is indeed valid. You don't even need the reverse triangle inequality. $$|s_n| = |s_n - s + s| \le |s_n -s | + |s| < |s| + 1$$ – EuYu Oct 15 '12 at 4:10
Is this true for all norms? – SPRajagopal Oct 8 '14 at 8:28
@SPRajagopal The only property we used in the proof was the triangle inequality itself, so this holds with any norm. – EuYu Oct 8 '14 at 14:05

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.

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