8. Given three positive numbers $a, b, c$ satisfying $a \leqslant b+c \leqslant 3 a, 3 b^{2} \leqslant a(a+c) \leqslant 5 b^{2}$, then the minimum value of $\frac{b-2 c}{a}$ is
8. $-\frac{18}{5}$ Detailed Explanation: The given inequality can be transformed into
$1 \leqslant \frac{b}{a}+\frac{c}{a} \leqslant 3,3\left(\frac{b}{a}\right)^{2} \leqslant 1+\frac{c}{a} \leqslant 5\left(\frac{b}{a}\right)^{2}$.
Let $\frac{b}{a}=x, \frac{c}{a}=y$, then we have
$$
\left\{\begin{array}{l}
1 \leqslant x+y \leqslant 3, \\
3 x^{2} \leqslant 1+y \leqslant 5 x^{2}, \\
x>0, y>0
\end{array}\right.
$$

It represents the shaded region $A B C D$ (as shown in the right figure).
Let $u=\frac{b-2 c}{a}=x-2 y$, then $y=\frac{1}{2}(x-u)$, which represents a moving line with a slope of $\frac{1}{2}$.
It is easy to see that when the moving line passes through point $D$, $u$ reaches its minimum value.
Solving the system of equations $\left\{\begin{array}{l}x+y=3, \\ 1+y=5 x^{2},\end{array}\right.$ we get the coordinates of point $D$ as $\left(\frac{4}{5}, \frac{11}{5}\right)$, so
$$
u_{\text {min }}=\frac{4}{5}-2 \times \frac{11}{5}=-\frac{18}{5} \text {. }
$$

Therefore, the minimum value of $\frac{b-2 c}{a}$ is $-\frac{18}{5}$.