metric spaces - basic inequality 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. 
 A: 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: 


*

*$d(x,z) \leq d(x,y) + d(y,z)$ 


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


*$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 $-d(x,y) \leq d(x,z) - d(y,z) \leq d(x,y) \Rightarrow d(x,y) \geq |d(x,z) - d(y,z)|$
A: 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,
$$
|d(\alpha,\theta)-d(\beta,\theta)|\le d(\alpha,\beta)
$$
