Example 5 Let $a, b, c$ be positive real numbers. Find the minimum value of
$$
\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}
$$

(2004, Girls' Mathematical Olympiad)
Let $a+2b+c=x, a+b+2c=y, a+b+3c=z$, then
$$
\begin{array}{l}
a+3c=2y-x, b=x+z-2y, c=z-y. \\
\text{Therefore, } \frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c} \\
=\frac{2y-x}{x}+\frac{4(x+z-2y)}{y}-\frac{8(z-y)}{z} \\
=-17+\frac{2y}{x}+\frac{4x}{y}+\frac{4z}{y}+\frac{8y}{z} \\
\geqslant-17+2\sqrt{8}+2\sqrt{32}=-17+12\sqrt{2}.
\end{array}
$$

The equality holds if and only if $\frac{2y}{x}=\frac{4x}{y}, \frac{4z}{y}=\frac{8y}{z}$, i.e., $y=\sqrt{2}x$, $z=2x$.
From $\left\{\begin{array}{l}a+b+2c=\sqrt{2}(a+2b+c), \text{ we get } \\ a+b+3c=2(a+2b+c)\end{array}\right.$
$$
\left\{\begin{array}{l}
b=(1+\sqrt{2})a, \\
c=(4+3\sqrt{2})a.
\end{array}\right.
$$

Therefore, for any positive real number $a$, when $b=(1+\sqrt{2})a$, $c=(4+3\sqrt{2})a$, the original expression achieves its minimum value of $-17+12\sqrt{2}$.