21
$\begingroup$

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

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

Why?

$\endgroup$
  • 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
41
$\begingroup$

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

$\endgroup$
  • 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
  • 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
  • 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
  • 1
    $\begingroup$ is there an intuitive explanation for why this is true? $\endgroup$ – Charlie Parker Nov 2 '17 at 2:37
  • 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
15
$\begingroup$

No. For example, $|(-2)-3|=5>|-2|-|3|=-1.$

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

$\endgroup$
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.