9. Given positive numbers $a, b, c, d$ satisfying $a+b+c+d=1$, find the minimum value of $M=\sqrt{a^{2}+\frac{1}{8 a}}+\sqrt{b^{2}+\frac{1}{8 b}}+\sqrt{c^{2}+\frac{1}{8 c}}+\sqrt{d^{2}+\frac{1}{8 d}}$.
$$
a^{2}+\frac{1}{8 a}=a^{2}-2 a+2 a+\frac{1}{8 a} \geqslant a^{2}-2 a+2 \sqrt{2 a \cdot \frac{1}{8 a}}=a^{2}-2 a+1=(a-1)^{2} \text {, }
$$

Then, $\sqrt{a^{2}+\frac{1}{8 a}} \geqslant 1-a$, with equality when $2 a=\frac{1}{8 a} \Rightarrow a=\frac{1}{4}$.
Thus, $\sqrt{a^{2}+\frac{1}{8 a}}+\sqrt{b^{2}+\frac{1}{8 b}}+\sqrt{c^{2}+\frac{1}{8 c}}+\sqrt{d^{2}+\frac{1}{8 d}} \geqslant 4-(a+b+c+d)=3$, with equality when $a=b=c=d=\frac{1}{4}$. Therefore, the minimum value of $M$ is 3.