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 $x=a+2 b+c, y=a+b+2 c, z=a+b+3 c$. Solving these equations simultaneously, we get $a+3 c=2 y-x$, $b=z+x-2 y, c=z-y$. Therefore, the expression we are looking for is
$$
\begin{array}{c}
\frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z}=-17+2 \frac{y}{x}+4 \frac{x}{y}+4 \frac{z}{y}+8 \frac{y}{z} \\
\geqslant-17+2 \sqrt{8}+2 \sqrt{32}=-17+12 \sqrt{2}
\end{array}
$$

It is easy to see that the equality holds when $y=\sqrt{2} x, z=2 x$. That is, when $b=(1+\sqrt{2}) a, c=(4+3 \sqrt{2}) a, a \in \mathbf{R}^{+}$, the minimum value is $-17+12 \sqrt{2}$.