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)
Let
\[
\left\{
\begin{array}{l}
x = a + 2b + c, \\
y = a + b + 2c, \\
z = a + b + 3c.
\end{array}
\right.
\]
Then we have \( x - y = b - c \) and \( z - y = c \), which leads to
\[
\left\{
\begin{array}{l}
a + 3c = 2y - x, \\
b = z + x - 2y, \\
c = z - y.
\end{array}
\right.
\]

Thus,
\[
\begin{array}{l}
\frac{a + 3c}{a + 2b + c} + \frac{4b}{a + b + 2c} - \frac{8c}{a + b + 3c} = \\
\frac{2y - x}{x} + \frac{4(z + x - 2y)}{y} - \frac{8(z - y)}{z} = \\
-17 + 2\left(\frac{y}{x} + \frac{2x}{y}\right) + 4\left(\frac{z}{y} + \frac{2y}{z}\right) \geq \\
-17 + 4 \sqrt{\frac{y}{x} \cdot \frac{2x}{y}} + 8 \sqrt{\frac{z}{y} \cdot \frac{2y}{z}} = -17 + 12 \sqrt{2}
\end{array}
\]

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

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 \(a\), if \(b = (1 + \sqrt{2})a\) and \(c = (4 + 3\sqrt{2})a\), then
\[
\frac{a + 3c}{a + 2b + c} + \frac{4b}{a + b + 2c} - \frac{8c}{a + b + 3c} = -17 + 12\sqrt{2}.
\]

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