2
$\begingroup$

The reverse Triangle Inequality states that $|a-b|\geq ||a|-|b||$ for any $a,b\in \mathbb R$. What about $$|a-b-c|\geq ||a|-|b|-|c|| \tag{*}$$ I know you will say its so elementary question, but I want to be sure:

So, repeating the original inequality for two numbers we get

$$|(a-b)-c|\geq \big||a-b|-|c|\big|\geq \bigg|\big||a|-|b|\big|-|c|\bigg|$$

should we have $|a|\geq |b|$ to get the required inequality in $(*)$?

$\endgroup$
2
  • $\begingroup$ In (*), what if $a=1$, $b=4$, $c=-4$? $\endgroup$ Jun 24 '12 at 23:56
  • $\begingroup$ Related $\endgroup$
    – leo
    Nov 4 '14 at 4:29
1
$\begingroup$

Your inequality is incorrect. Let $a=2$, $b=-1$, and $c=3$. Then,

$|a-b-c|=0$, but

$||a|-|b|-|c||=|2-1-3|=2$. So $|a-b-c|<||a|-|b|-|c||$ in this counterexample.

The first mistake in the proof presented in the question is in the step

$||a-b|-|c||\ge|||a|-|b||-|c||$

Notice that this implicitly assumes that if $a\ge b$, then $|a-c|\ge|b-c|$, which is false in general if $a>c>b$.

$\endgroup$
4
  • $\begingroup$ So what is the correct inequality for $|a-b-c|$ ? $\endgroup$
    – Simplemath
    Jun 25 '12 at 0:01
  • 1
    $\begingroup$ Besides $|a-b-c|\ge|a|-|b|-|c|$, I do not think that one exists. $\endgroup$
    – A.S
    Jun 25 '12 at 0:05
  • $\begingroup$ And this will be true no matter which one is smaller or bigger than others! $\endgroup$
    – Simplemath
    Jun 25 '12 at 0:07
  • $\begingroup$ Actually, I'm looking for some positive lower bound for $|a-b-c|$? $\endgroup$
    – Simplemath
    Jun 25 '12 at 0:17

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.