4. Let $a, b$, and $c$ be three distinct real numbers. Find the minimum possible value of the expression

$$
\left|\frac{a}{b-c}\right|+\left|\frac{b}{c-a}\right|+\left|\frac{c}{a-b}\right|
$$

if it exists, as well as all triples $(a, b, c)$ for which this value is achieved.
4. If we denote $x=\frac{a}{b-c}, y=\frac{b}{c-a}$ and $z=\frac{c}{a-b}$, we have $x-1=\frac{a-b+c}{b-c}, x+1=\frac{a+b-c}{b-c}$, etc., from which we conclude that

$$
(x-1)(y-1)(z-1)=(x+1)(y+1)(z+1), \quad \text { i.e. } \quad x y+y z+z x=-1
$$

Then $|x y|+|y z|+|z x| \geqslant 1$, from which it follows that

$$
(|x|+|y|+|z|)^{2}=(x+y+z)^{2}-2(x y+y z+z x)+2(|x y|+|y z|+|z x|) \geqslant 0+2+2=4
$$

i.e. $|x|+|y|+|z| \geqslant 2$.

Equality holds only if $x+y+z=0$. Since $|x|+|y|+|z|= \pm x \pm y \pm z=2$, adding them together implies that one of the numbers $x, y, z$, say $z$, is equal to $\pm 1$.

- If $z=1$, this means $a=b+c$ and from there $y=-1$, so it must be $x=0$, i.e. $a=0$.
- Similarly, if $z=-1$, we get $b=a+c, x=1$ and $y=0$, i.e. $b=0$.

Therefore, the minimum value of the given expression is 2 and it is achieved when one of the numbers is zero and the sum of the remaining two is zero.