$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ 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.

Notice that $|x-y|\leq |x|+|y|$ implies that:
$$\frac{|z|}{|x-y|} \geq \frac{|z|}{|x|+|y|}.$$
Then:
$$S \geq \frac{|a|}{|b|+|c|} + \frac{|b|}{|a|+|c|} + \frac{|c|}{|a|+|b|} = \\
=\frac{A}{B+C} + \frac{B}{A+C} + \frac{C}{A+B},$$
 where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S \geq \frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S \geq \frac{D + A^3 +B^3 + C^3 + ABC}{D} > \frac{D}{D} = 1.$$
