Example 11 Given that $a, b, c$ are positive numbers, $a+b+c=10$, and $a \leqslant 2b, b \leqslant 2c, c \leqslant 2a$, find the minimum value of $abc$.

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Let $\quad x=2 b-a, y=2 c-b, z=2 a-c$,
then $\quad x+y+z=10, x \geqslant 0, y \geqslant 0, z \geqslant 0$,
and $\quad a=\frac{x+2 y+4 z}{7}, b=\frac{y+2 z+4 x}{7}, c=\frac{z+2 x+4 y}{7}$,
thus $a b c=\frac{1}{343} \cdot[(x+2 y+4 z)(y+2 z+4 x)(z+2 x+4 y)]$,

and
$$\begin{aligned}
& (x+2 y+4 z)(y+2 z+4 x)(z+2 x+4 y) \\
= & (10+y+3 z)(10+z+3 x)(10+x+3 y) \\
= & 1000+400(x+y+z)+130(x y+y z+z x)+30\left(x^{2}+y^{2}+z^{2}\right)+\left(3 y^{2} z\right. \\
& \left.+3 x^{2} y+9 x y^{2}+3 x z^{2}+9 y z^{2}+9 x^{2} z+28 x y z\right) \\
= & 1000+4000+30(x+y+z)^{2}+\left[70(x y+y z+z x)+3\left(y^{2} z+x^{2} y+\right.\right. \\
& \left.\left.x z^{2}\right)+9\left(x y^{2}+y z^{2}+x^{2} z\right)+28 x y z\right] \\
\geqslant & 8000 .
\end{aligned}$$

Equality holds if and only if $x=y=0, z=10$, at which point $a=\frac{40}{7}, b=\frac{20}{7}, c=\frac{10}{7}$.
Therefore, $a b c_{\min }=\frac{8000}{343}$.