10 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 $\qquad$
(10) Given the inequality can be transformed into
$$
\begin{array}{r}
1 \leqslant \frac{b}{a}+\frac{c}{a} \leqslant 3, \\
3\left(\frac{b}{a}\right)^{2} \leqslant 1+\frac{c}{a} \leqslant 5\left(\frac{b}{a}\right)^{2} .
\end{array}
$$

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 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 can find the coordinates of point $D$ to be $\left(\frac{4}{5}, \frac{11}{5}\right)$, so
$$
u_{\text {min }}=\frac{4}{5}-2 \times \frac{11}{5}=-\frac{18}{5} .
$$

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