Example 5 Let $a, b, c$ be positive real numbers, and $abc + a + c = b$. Try to 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)
Analysis: From the given conditions, we have $a+c=(1-a c) b$. Clearly, $1-a c \neq 0$. Therefore, $b=\frac{a+c}{1-a c}$. Let $a=\tan \alpha$, $b=\tan \beta, c=\tan \gamma, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$. Then
$$
\tan \beta=\frac{\tan \alpha+\tan \gamma}{1-\tan \alpha \cdot \tan \gamma}=\tan (\alpha+\gamma) .
$$

Since $\beta, \alpha+\gamma \in(0, \pi)$, it follows that $\beta=\alpha+\gamma$. Thus,
$$
\begin{array}{l}
p=\frac{2}{1+\tan ^{2} \alpha}-\frac{2}{1+\tan ^{2} \beta}+\frac{3}{1+\tan ^{2} \gamma} \\
=2 \cos ^{2} \alpha-2 \cos ^{2}(\alpha+\gamma)+3 \cos ^{2} \gamma \\
=(1+\cos 2 \alpha)-[1+\cos (2 \alpha+2 \gamma)]+3 \cos ^{2} \gamma \\
=2 \sin \gamma \cdot \sin (2 \alpha+\gamma)+3 \cos ^{2} \gamma \\
\leqslant 2 \sin \gamma+3\left(1-\sin ^{2} \gamma\right) \\
=-3\left(\sin \gamma-\frac{1}{3}\right)^{2}+\frac{10}{3} \leqslant \frac{10}{3},
\end{array}
$$

Equality holds if and only if $2 \alpha+\gamma=\frac{\pi}{2}$ and $\sin \gamma=\frac{1}{3}$. Therefore, when $a=\frac{\sqrt{2}}{2}, b=\sqrt{2}, c=\frac{\sqrt{2}}{4}$, $p_{\max }=\frac{10}{3}$.

Note: Expressions of the form $\frac{2 x}{1 \pm x^{2}}, \frac{1-x^{2}}{1+x^{2}}, \frac{x-y}{1+x y}, \frac{x+y}{1-x y}$, $\sqrt{a^{2}+x^{2}}$ can use tangent (or cotangent) substitutions for $x, y$.