$$
\begin{array}{l}
\text { 10. Let } S=x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+ \\
x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2} . \\
\text { Given }\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\right) \\
\leqslant \frac{\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)^{2}}{27}=\frac{1}{27} \text {, } \\
\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\right) \\
=x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6}+S \geqslant \frac{1}{540}+S,
\end{array}
$$

then $S \leqslant \frac{1}{27}-\frac{1}{540}=\frac{19}{540}$.
When $x_{1}=0, x_{2}=\frac{1}{60}, x_{3}=0, x_{4}=\frac{1}{3}, x_{5}=\frac{19}{60}$, $x_{6}=\frac{1}{3}$, the equality holds.
$$
10. (20 points) Let $x_{i} \geqslant 0(i=1,2, \cdots, 6)$, and satisfy
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{6}=1, \\
x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geqslant \frac{1}{540} .
\end{array}\right.
$$

Find the maximum value of $x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+$ $x_{6} x_{1} x_{2}$.