Why is $|a - b| \geq|a| - |b|$?
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asked Oct 15 '12 at 3:56
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1$\begingroup$ it would be also nice to see an intuitive explanation/answer to this. It feels it should be obvious, no? $\endgroup$ – Charlie Parker Nov 2 '17 at 2:40
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$\begingroup$ Does this answer your question? Reverse Triangle Inequality Proof $\endgroup$ – Chris Tang Feb 25 '20 at 7:14
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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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1$\begingroup$ Let $s_n$ be a sequence. Is this valid then: $|s_n - s| < 1 \iff ||s_n| - |s|| < 1 \iff |s_n| < |s| + 1$? $\endgroup$ – CodeKingPlusPlus Oct 15 '12 at 4:08
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1$\begingroup$ That is indeed valid. You don't even need the reverse triangle inequality. $$|s_n| = |s_n - s + s| \le |s_n -s | + |s| < |s| + 1$$ $\endgroup$ – EuYu Oct 15 '12 at 4:10
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2$\begingroup$ @SPRajagopal The only property we used in the proof was the triangle inequality itself, so this holds with any norm. $\endgroup$ – EuYu Oct 8 '14 at 14:05
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1$\begingroup$ is there an intuitive explanation for why this is true? $\endgroup$ – Charlie Parker Nov 2 '17 at 2:37
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1$\begingroup$ @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. $\endgroup$ – EuYu Nov 2 '17 at 13:10
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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.
answered Oct 15 '12 at 4:14
