Example 1 Solve the system of equations
$$\left\{\begin{array}{l}
a^{2}=\frac{\sqrt{b c} \sqrt[3]{b c d}}{(b+c)(b+c+d)} \\
b^{2}=\frac{\sqrt{c d} \sqrt[3]{c d a}}{(c+d)(c+d+a)} \\
c^{2}=\frac{\sqrt{d a} \sqrt[3]{d a b}}{(d+a)(d+a+b)} \\
d^{2}=\frac{\sqrt{a b} \sqrt[3]{a b c}}{(a+b)(a+b+c)}
\end{array}\right.$$

for real solutions.
First, note that none of the variables equals zero. Without loss of generality, assume $b=0$, from (31) we get $a=0$, from (34) we get $d=0$, and from (33) we get $c=0$, which means all values are zero, but this is impossible because the denominator would be zero.

Second, note that the square roots of $b c, c d, d a, a b$ must all exist, which means $a, b, c, d$ must all be negative or all be positive. If they are all negative, the right-hand side of these equations would be negative, contradicting the fact that they are squares of real numbers. Therefore, the four values must all be positive.

According to the arithmetic-geometric mean inequality, we have
$$\sqrt{b c} \leqslant \frac{b+c}{2} \text {, i.e., } \frac{\sqrt{b c}}{b+c} \leqslant \frac{1}{2} \text { and } \sqrt[3]{b c d} \leqslant \frac{b+c+d}{3},$$

i.e., $\square$
$$\frac{\sqrt[3]{b c d}}{b+c+d} \leqslant \frac{1}{3}$$

Therefore, $a^{2}=\frac{\sqrt{b c} \sqrt[3]{b c d}}{(b+c)(b+c+d)} \leqslant \frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$.
Thus, $a \leqslant \frac{1}{\sqrt{6}}$.
Similarly, we have $b \leqslant \frac{1}{\sqrt{6}}, c \leqslant \frac{1}{\sqrt{6}}, d \leqslant \frac{1}{\sqrt{6}}$.
From this, we get $(b+c)(b+c+d) \leqslant \frac{2}{\sqrt{6}} \times \frac{3}{\sqrt{6}}=1$.
Similarly, we have
$$\begin{array}{l}
(c+d)(c+d+a) \leqslant 1 \\
(d+a)(d+a+b) \leqslant 1 \\
(a+b)(a+b+c) \leqslant 1
\end{array}$$

From $(31) \times(32) \times(33) \times(34)$, we get
$$1=(b+c)(b+c+d)(c+d)(c+d+a) \cdot(d+a)(d+a+b)(a+b)(a+b+c)$$

Since the product of 4 expressions that are each less than or equal to 1 equals 1, these 4 expressions must all equal 1.

Thus, the only possibility is that each variable takes its maximum possible value.
Therefore, $a=b=c=d=\frac{\sqrt{6}}{6}$ is the unique solution to the given system of equations.