9. Let $a, b, c$ be positive real numbers, and $a b c + a + c = b$. Determine the maximum value of
$$
p=\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}+\frac{3}{c^{2}+1}
$$

(1999, Vietnam Mathematical Olympiad)
Given the condition $a+c=(1-a c) b$, it is clear that $1-a c \neq 0$, hence $b=\frac{a+c}{1-a c}$.

Let $\alpha=\arctan a, \beta=\arctan b, \gamma=\arctan c, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, then
$$
\tan \beta=\frac{\tan \alpha+\tan \gamma}{1-\tan \alpha \cdot \tan \gamma}=\tan (\alpha+\gamma).
$$

Since $\beta, \alpha+\gamma \in(0, \pi)$, we have $\beta=\alpha+\gamma$. Therefore,
$$
\begin{aligned}
p= & \frac{2}{1+\tan ^{2} \alpha}-\frac{2}{1+\tan ^{2} \beta}+\frac{3}{1+\tan ^{2} \gamma} \\
= & 2 \cos ^{2} \alpha-2 \cos ^{2}(\alpha+\gamma)+3 \cos ^{2} \gamma \\
& =(\cos 2 \alpha+1)-[\cos (2 \alpha+2 \gamma)+1]+3 \cos ^{2} \gamma \\
& =2 \sin \gamma \cdot \sin (2 \alpha+\gamma)+3 \cos ^{2} \gamma \\
& \leqslant 2 \sin \gamma+3 \cos ^{2} \gamma \\
& =\frac{10}{3}-3\left(\sin \gamma-\frac{1}{3}\right)^{2} \leqslant \frac{10}{3},
\end{aligned}
$$

and the equality holds when $2 \alpha+\gamma=\frac{\pi}{2}, \sin \gamma=\frac{1}{3}$, i.e., $a=\frac{\sqrt{2}}{2}, b=\sqrt{2}, c=\frac{\sqrt{2}}{4}$. Hence, $p_{\text {max }}=\frac{10}{3}$.