<s>[INST] $\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ 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.
 [/INST] Let $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=\sum_{cyc}\frac{ab}{(b-c)(c-a)}=\frac{\sum\limits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$\sum_{cyc}\left|\frac{a}{b-c}\right|=\sqrt{\left(|x|+|y|+|z|\right)^2}=\sqrt{x^2+y^2+z^2+2\sum\limits_{cyc}|xy|}=$$
$$=\sqrt{(x+y+z)^2+2+2\sum\limits_{cyc}|xy|}\geq\sqrt{2+2}=2.$$
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