Example 5 Let $a, b, c$ be positive real numbers, find the minimum value of
$$\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}$$
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 $x-y=b-c, z-y=c$, which leads to
$$\left\{\begin{array}{l}
a+3 c=2 y-x \\
b=z+x-2 y \\
c=z-y
\end{array}\right.$$

Thus,
$$\begin{aligned}
& \frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c} \\
= & \frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z} \\
= & -17+2 \frac{y}{x}+4 \frac{x}{y}+4 \frac{z}{y}+8 \frac{y}{z} \\
\geqslant & -17+2 \sqrt{8}+2 \sqrt{32}=-17+12 \sqrt{2}
\end{aligned}$$

The equality in the above expression can hold. In fact, from the derivation process, the equality holds if and only if the equality in the arithmetic mean inequality holds, which is equivalent to
$$\left\{\begin{array} { l } 
{ 2 \frac { y } { x } = 4 \frac { x } { y } , } \\
{ 4 \frac { z } { y } = 8 \frac { y } { z } , }
\end{array} \text { i.e., } \left\{\begin{array} { l } 
{ y ^ { 2 } = 2 x ^ { 2 } , } \\
{ z ^ { 2 } = 2 y ^ { 2 } , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
y=\sqrt{2} x \\
z=2 x
\end{array}\right.\right.\right.$$

That is,
$$\left\{\begin{array}{l}
a+b+2 c=\sqrt{2}(a+2 b+c) \\
a+b+3 c=2(a+2 b+c)
\end{array}\right.$$

Solving this 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 positive 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 $-17+12 \sqrt{2}$.