2. Let $a$, $b$, $c$ be positive real numbers. Find the minimum value of
$$
\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}
$$

(Li Shenghong)
2. $-17+12 \sqrt{2}$.

Let $x=a+2 b+c, y=a+b+2 c, z=a+b+$
$3 c$, then we have $x-y=b-c, z-y=c$. Therefore, we get
$$
\begin{array}{l}
a+3 c=2 y-x, b=z+x-2 y, c=z-y . \\
\text { Hence } \frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c} \\
=\frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z} \\
=-17+2 \frac{y}{x}+4 \frac{x}{y}+4 \frac{z}{y}+8 \frac{y}{z} \\
\geqslant-17+2 \sqrt{8}+2 \sqrt{32}=-17+12 \sqrt{2} .
\end{array}
$$

The equality in the above expression can hold.
In fact, from the above 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 indeterminate equation, we get
$$
\left\{\begin{array}{l}
b=(1+\sqrt{2}) a, \\
c=(4+3 \sqrt{2}) a .
\end{array}\right.
$$

It is easy to calculate that for any positive real number $a$, as long as $b=(1+\sqrt{2}) a$, $c=(4+3 \sqrt{2}) a$, the equality in equation (1) holds.
Therefore, the minimum value sought is $-17+12 \sqrt{2}$.