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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 $(*)$?

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    In (*), what if $a=1$, $b=4$, $c=-4$?2012-06-24
  • 0
    [Related](http://math.stackexchange.com/q/734106/8271)2014-11-04

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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 $a>c>b$.

  • 0
    So what is the correct inequality for $|a-b-c|$ ?2012-06-25
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    Besides $|a-b-c|\ge|a|-|b|-|c|$, I do not think that one exists.2012-06-25
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    And this will be true no matter which one is smaller or bigger than others!2012-06-25
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    Actually, I'm looking for some positive lower bound for $|a-b-c|$?2012-06-25