Find the minimum value of $x+y+z.$ Let $x,y,z$ are nonegative  such that $(x - y)(y - z)(z - x) \geq 1.$ 
Find the minimum value  of  $x+y+z.$ 

Clearly, two among $x-y, y-z, z-x$ should be negative. WLOG let $a=x-y, b=z-y$ be positive numbers. Then $a(-b)(b-a) \ge 1 \iff a^2b \ge ab^2+1$ and we need the minimum of $a+b+3y$.    Obviously it is a good idea to set $y =0$, then we are left with minimising $a+b$.
$$a^2b-b^2a-1 \ge 0 \iff a \ge \frac{b^2+\sqrt{b^4+4b}}{2b}$$
So we minimise $a+b =\frac32b+ \dfrac{\sqrt{b^4+4b}}{2b}$, which by calculus and some algebra gives $b = \sqrt[3]{\frac{3\sqrt3-5}2}$.
Therefore we have the minimum (after some more algebra and help from @Semiclassical) as $\sqrt[3]2 \sqrt3$.  
