Example 7 Given that $a, b, c, d$ are positive real numbers. Find
\[
\begin{aligned}
S= & \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b} \\
& +\frac{d}{a+b+c}
\end{aligned}
\]

the minimum value.
Solution: Let $b+c+d=A, c+d+a=B, d+$ $a+b=C, a+b+c=D$.
Adding the above four equations, we get
$$
a+b+c+d=\frac{1}{3}(A+B+C+D) .
$$

Subtracting the four equations from equation (1) respectively, we get
$$
\begin{array}{c}
a=\frac{1}{3}(B+C+D-2 A), \\
b=\frac{1}{3}(C+D+A-2 B), \\
c=\frac{1}{3}(D+A+B-2 C), \\
d=\frac{1}{3}(A+B+C-2 D) .
\end{array}
$$

The original expression can be transformed into
$$
\begin{aligned}
S= & \frac{B+C+D-2 A}{3 A}+\frac{C+D+A-2 B}{3 B} \\
& +\frac{D+A+B-2 C}{3 C}+\frac{A+B+C-2 D}{3 D},
\end{aligned}
$$

i.e., $3 S+12$
$$
\begin{aligned}
= & \frac{A+B+C+D}{A}+\frac{A+B+C+D}{B} \\
& +\frac{A+B+C+D}{C}+\frac{A+B+C+D}{D} \\
= & (A+B+C+D) \cdot\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D}\right) \\
\geqslant & 16,
\end{aligned}
$$

i.e., $3 S+12 \geqslant 16, S \geqslant \frac{4}{3}$.
Equality holds if and only if $A=B=C=D$ (i.e., $a=b=c=d$).
Therefore, the minimum value of $S$ is $\frac{4}{3}$.