$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that  $  S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.

Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|\le |c|+|a|$ and $|a-b| \le |a|+|b-c|+|c|$. Therefore
$$
\frac{|a|}{|b-c|}+\frac{|b|}{|c-a|}+\frac{|c|}{|a-b|} \ge \frac{|a|}{|b-c|} + \frac{|b-c|+|c|}{|c|+|a|} + \frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 \ge 2xy$, we get
\begin{align*}
(*) & = \frac{x}{y}+\frac{y+z}{z+x}+\frac{z}{x+y+z} \\
&\ge \frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \\
&= \frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \\
&\ge \frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \\
&=2.\end{align*}
