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 $(*)$?

  • $\begingroup$ 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

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


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

  • $\begingroup$ 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

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