Example 10 Let $a, b, c$ be positive real numbers, find the minimum value of $\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}$. (3rd China Girls Mathematical Olympiad problem)
Let
$$\left\{\begin{array}{l}x=a+2 b+c, \\ y=a+b+2 c \\ z=a+b+3 c .\end{array}\right.$$
Then we have $x-y=b-c, z-y=c$, from which we can derive
$$\left\{\begin{array}{l}
a+3 c=2 y-x \\
b=z+x-2 y \\
c=z-y
\end{array}\right.$$

Thus,
$$\begin{aligned}
& \frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c} \\
= & \frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z} \\
= & -17+2\left(\frac{y}{x}+\frac{2 x}{y}\right)+4\left(\frac{z}{y}+\frac{2 y}{z}\right) \\
\geqslant & -17+4 \sqrt{\frac{y}{x} \cdot \frac{2 x}{y}}+8 \sqrt{\frac{z}{y} \cdot \frac{2 y}{z}}=-17+12 \sqrt{2} .
\end{aligned}$$

The equality in the above expression can be achieved. In fact, from the above derivation, the equality holds if and only if the equality in the AM-GM inequality holds, which is equivalent to
$$\left\{\begin{array}{l}\frac{y}{x}=\frac{2 x}{y}, \\ \frac{z}{y}=\frac{2 y}{z},\end{array}\right.$$
i.e.,
$$\left\{\begin{array}{l}y^{2}=2 x^{2} \\ z^{2}=2 y^{2}\end{array}\right.,$$
which is also equivalent to
$$\left\{\begin{array}{l}
y=\sqrt{2} x, \\ z=2 y^{2},
\end{array}\right.$$
or
$$\left\{\begin{array}{l}
a+b+2 c=\sqrt{2}(a+2 b+c) \\
a+b+3 c=\sqrt{2}(a+b+2 c)
\end{array}\right.$$

Solving this system of equations, we get
$$\left\{\begin{array}{l}
b=(1+\sqrt{2}) a \\
c=(4+3 \sqrt{2}) a
\end{array}\right.$$

It is not difficult to verify that for any real number $a$, as long as $b=(1+\sqrt{2}) a, c=(4+3 \sqrt{2}) a$, we have
$$\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}=-17+12 \sqrt{2},$$

Therefore, the minimum value sought is $-17+12 \sqrt{2}$.