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Why is $|a - b| \geq|a| - |b|$?

lithium123
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CodeKingPlusPlus
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3 Answers3

67

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 $$-|a-b|\le|a|-|b| \le |a - b|$$

EuYu
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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 04:08
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    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 04:10
  • Is this true for all norms? – sprajagopal Oct 08 '14 at 08:28
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    @SPRajagopal The only property we used in the proof was the triangle inequality itself, so this holds with any norm. – EuYu Oct 08 '14 at 14:05
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    is there an intuitive explanation for why this is true? – Charlie Parker Nov 02 '17 at 02:37
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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. – EuYu Nov 02 '17 at 13:10
  • Finally a good answer with a good proof, you'd think it would be more common on this site. Thank you! – jeffery_the_wind Oct 12 '19 at 16:56
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No. For example, $|(-2)-3|=5>|-2|-|3|=-1.$

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

Andrew
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10

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.

Salech Alhasov
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