Minimising expression with absolute values

The expression for $L$ does not change if $x, y,z$ are permuted, so you may indeed assume that $x > y > z$.

With $a = x-y > 0$ and $b = y-z > 0$ we have $$ |x| + |y| + |z| \ge x-z = a+b $$ and $$ |(x-y)(x-z)(y-z)| = a(a+b)b $$ and therefore $$ L \ge \frac{(a+b)^3}{a(a+b)b} = \frac{(a+b)^2}{ab} = \frac{(a-b)^2}{ab} + 4 \ge 4\, . $$

Equality holds exactly if $y=0$ and $a=b$, i.e. for $(x,y,z) = (a, 0, -a)$ with $a > 0$.