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Let $(\Omega, d)$ be a metric space. I have to show that $ d(\alpha ,\beta) \ge | d(\alpha, \theta) - d(\theta, \beta)|$ for every $\alpha, \beta, \theta \in \Omega.$

Starting with the triangle inequality does not help much. $ d(\alpha, \beta) \le d(\alpha, \theta) + d(\theta, \beta) $. The only somewhat logical way is to assume that $\min_{} \{ d(\alpha, \theta), d(\theta, \beta) \} \le \max_{} \{ d(\alpha, \theta), d(\theta, \beta) \} \le d(\alpha, \beta)$ and then proceed by proving that $\max$ is a metric. But, it seems too complicated for this kind of question and implies pretty unnecessary assumptions.

Will appreciate any help.

Guest1
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2 Answers2

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This is standard proof of reverse triangle inequality on metric spaces

Let $x = \alpha, z = \beta, y = \theta$

By triangle inequality:

$d(x,z) \leq d(x,y) + d(y,z) \Rightarrow d(x,z) - d(y,z) \leq d(x,y)$

Case 1: If $d(x,z) - d(y,z) \geq 0$, then $d(x,y) \geq |d(x,z) - d(y,z)|$

Case 2:

  1. $d(x,z) \leq d(x,y) + d(y,z)$

$\Rightarrow d(x,z) - d(y,z) \leq d(x,y)$

  1. $d(y,z) \leq d(z,x) + d(x,y) \Rightarrow d(y,z) - d(x,z) \leq d(x,y) $

$\Rightarrow d(x,z) - d(y,z) \geq -d(x,y)$

Thus $-d(x,y) \leq d(x,z) - d(y,z) \leq d(x,y) \Rightarrow d(x,y) \geq |d(x,z) - d(y,z)|$

Fraïssé
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You're starting with the wrong version of the triangle inequality; start from $$ d(\alpha,\theta)\le d(\alpha,\beta)+d(\beta,\theta) $$ which implies $$ d(\alpha,\theta)-d(\theta,\beta)\le d(\alpha,\beta)\tag{1} $$ Similarly, from $$ d(\beta,\theta)\le d(\beta,\alpha)+d(\alpha,\theta) $$ we deduce $$ -d(\alpha,\beta)\le d(\alpha,\theta)-d(\beta,\theta)\tag{2} $$ Now recall that, if $r\ge0$, $-r\le s\le r$ is the same as $|s|\le r$, so from $(1)$ and $(2)$ we deduce $$ -d(\alpha,\beta)\le d(\alpha,\theta)-d(\beta,\theta)\le d(\alpha,\beta) $$ that is, $$ |d(\alpha,\theta)-d(\beta,\theta)|\le d(\alpha,\beta) $$

egreg
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