Example 8 (2004 National Girls' Mathematical Olympiad) Let $a, b, c$ be positive real numbers. Find the minimum value of $\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-$ $\frac{8 c}{a+b+3 c}$.
This is a problem of finding the extremum of a ternary function in the form of fractional algebra, where the denominators of the three fractions are relatively complex. For simplicity, let's denote them with a single letter, i.e., 
let $a+2 b+c=x, a+b+2 c=y, a+b+3 c=z$, then we have $x-y=b-c, z-y=c$. From this, we get $a+3 c=2 y-x, b=z+x-2 y, c=z-y$.
Thus, $\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}$
$=\frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z}$
$=-17+2 \frac{y}{x}+4 \frac{x}{y}+4 \frac{z}{y}+8 \frac{y}{z}$
$\geqslant-17+2 \sqrt{8}+2 \sqrt{32}=-17+12 \sqrt{2}$.
The equality in the above expression can hold.
In fact, from the above derivation, the equality holds if and only if the equality in the arithmetic mean inequality holds, which is equivalent to
$\left\{\begin{array}{l}2 \frac{y}{x}=4 \frac{x}{y}, \\ 4 \frac{z}{y}=8 \frac{y}{z},\end{array}\right.$
i.e., $\left\{\begin{array}{l}y^{2}=2 x^{2} ， \\ z^{2}=2 y^{2} ，\end{array}\right.$
i.e., $\left\{\begin{array}{l}y=\sqrt{2} x, \\ z=2 x,\end{array}\right.$
which is also equivalent to $\left\{\begin{array}{l}a+b+2 c=\sqrt{2}(a+2 b+c), \\ a+b+3 c=2(a+2 b+c) .\end{array}\right.$
Solving this indeterminate equation, we get
$$
\left\{\begin{array}{l}
b=(1+\sqrt{2}) a \\
c=(4+3 \sqrt{2}) a .
\end{array}\right.
$$