About $−\vert x \vert\le x \le \vert x \vert$ in absolute values So, i'm really strugling with this one, when studying the triangle inequality, the inequality  $−|x|≤x≤|x|$ pops really often, however I just can't get why this is true.  I've seen only two different approaches, one that it seems to me too trivial: 
Since:
$|x|≤a⇔−a≤x≤a$
We can write:
$|x|≤|x|$
as
$−|x|≤x≤|x|$
Is that reasoning even valid?
And the other way, very informal, is that because $x=|x|$  or $x=−|x|$  depending on the sign of $x$ then the inequality is satisfied. 
Why does the distance $x$ has the possibility to be greater than $x$ itself, shouldn't it only be that $x=|x|$ in any case? why greater than?
I just don't get it, and if I substitute in the inequality for $x>0$ and $x<0$ I don't have any idea what to do with it:
For $x<0$:
$ −(−x)≤x≤−x$ 
that is
$x≤x≤−x$
For $x>0$:
$x≤x≤x$
I know this should be really easy to grasp, but I just can't really REALLY understand how to use that inequality.
 A: *

*If $x = 0$ then it's obvious that $-|x| \leq x \leq |x|$ because $$-|x| = x = |x| = 0.$$

*If $x > 0$ then $x = |x|$ and
$$ -\underbrace{x}_{=|x|} < 0 < x =|x|.$$
Therefore $-|x| \leq x \leq |x|.$

*If $x < 0$ then $x = -|x|$ and
$$  -|x| = x< 0 < -\underbrace{x}_{-|x|} = |x|.$$
Therefore $-|x| \leq x \leq |x|.$

Why does the distance $x$ has the possibility to be greater than $x$ itself, shouldn't it only be that $x=|x|$ in any case? why greater than?

If $x = y$ than it is also true that $x \leq y$. The latter statement is less precise than the former, sure, but it is a true statement.
At least one of these inequalities 
$$-|x| \leq x \leq |x|$$
is actually an equality. Which one of them ? Well it depends on the nature of $x$ (positive, negative, zero). Therefore it's better to put inequalities because it is true for all $x$.

I've seen only two different approaches, one that it seems to me too trivial [...] Is that reasoning even valid?

Yes, it is.
A: The value of $x$ must be either $-|x|$ or $|x|$ depending on the sign of $x$. This can be expressed as the inequality you listed $-|x|\leq x\leq|x|$. I think you are trying to compare it to an inequality like $0\leq x\leq4$ where $x$ could take any value from $0$ to $4$. In your situation this is not the case, $x$ can only be one of two specific values, namely $-|x|$ or $|x|$.

Why does the distance $x$ has the possibility to be greater than $x$ itself, shouldn't it only be that $x=|x|$ in any case? why greater than?

The distance $|x|$ will always be a non-negative number where as $x$ can be a negative number so in some cases (i.e. $x<0$) we have that the distance is greater than the value (where greater is meaning signed magnitude not just magnitude).
