<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 is the standard technique for this. This one is a bit tricky, but anyways. Let WLOG $x\le y\le z$ and let $y=a+x,z=b+x$. Now plug this into 
$$(x+y+z)\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\ge C|(x-y)(y-z)(z-x)|$$
$$\iff(3x+a+b)(a^2-ab+b^2)\ge Cab(a-b)\quad \text{since } a\ge b$$
Now this is linear in $x$. Hence we may take $x=0$ to achieve the minimum of LHS. This gives
$$(a^3+b^3)\ge Cab(a-b)$$ We take $a=tb$ with $t\ge 1$.
Then $\displaystyle C_{\max}=\min_{t\ge 1}\dfrac{t^3+1}{t(t-1)}$
Which I believe is the one mentioned above. Is this working?
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