<s>[INST] How prove $x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$ 
let $x,y,z\ge 0$,and such 
  $$x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$$

Find the maximum of the $C$
witout loss of we assume that $$x+y+z=1$$
I think 
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)=1-3(yz+xz+xy)$$
then $$(x-y)(y-z)(x-z)=?$$
so I can't，It is said $$C_{max}=\sqrt{9+6\sqrt{3}}$$
 [/INST] this might be helpful
$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$
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