6. (50 points) Given that for any $x, y, z \geqslant 0$ there is $x^{3}+y^{3}+z^{3}-3 x y z \geqslant c|(x-y)(y-z)(z-x)|$. Find the maximum value of $c$.

保留源文本的换行和格式，直接输出翻译结果如下：

6. (50 points) Given that for any $x, y, z \geqslant 0$ there is $x^{3}+y^{3}+z^{3}-3 x y z \geqslant c|(x-y)(y-z)(z-x)|$. Find the maximum value of $c$.
6. The left side of the inequality
$$
\begin{array}{l}
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) \\
=\frac{(x+y+z)\left[(x-z)^{2}+(y-z)^{2}+(z-x)^{2}\right]}{2} .
\end{array}
$$

Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$.
If $x=y$ or $y=z$, the inequality holds for any real number $c$.

If $x>y>z \geqslant 0$, fix $x-y$ and $y-z$, then $z-x$ is also fixed.

To maximize $c$, we need to minimize the left side, i.e., minimize $x+y+z$.

Since $x+y+z=(x-y)+2(y-z)+3z$, the left side is minimized when $z=0$.
Now consider the case $x>y>z=0$.
The original expression becomes $x^{3}+y^{3} \geqslant c x y(x-y)$.
Let $t=\frac{x}{y}(t>1)$. Then $t^{3}+1 \geqslant c t(t-1)$.
Thus, $c_{\max }=f(t)_{\text {min }}$, where $f(t)=\frac{t^{3}+1}{t(t-1)}$.
Suppose $f(t)$ attains its minimum at $t_{0}$. Then
$$
\begin{array}{l}
f^{\prime}\left(t_{0}\right)=\frac{t_{0}^{4}-2 t_{0}^{3}-2 t_{0}+1}{t_{0}^{2}\left(t_{0}-1\right)^{2}}=0 . \\
\text { Hence }\left(t_{0}+\frac{1}{t_{0}}\right)^{2}-2\left(t_{0}+\frac{1}{t_{0}}\right)-2=0 .
\end{array}
$$

Solving this, we get $t_{0}=\frac{1+\sqrt{3}+\sqrt{2 \sqrt{3}}}{2}$.
Thus, $c_{\text {max }}=f\left(t_{0}\right)=\left(\frac{\sqrt{6}+3 \sqrt{2}}{2}\right) \sqrt[4]{3}$.