Example 2 Given real numbers $x, y, z > 3$, find all real solutions $(x, y, z)$ of the equation
$$\frac{(x+2)^{2}}{y+z-2}+\frac{(y+4)^{2}}{z+x-4}+\frac{(z+6)^{2}}{x+y-6}=36$$
Given $x, y, z > 3$, we know
$$y+z-2>0, z+x-4>0, x+y-6>0$$

By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& {\left[\frac{(x+2)^{2}}{y+z-2}+\frac{(y+4)^{2}}{x+z-4}+\frac{(z+6)^{2}}{x+y-6}\right] } \\
& {[(y+z-2)+(x+z-4)+(x+y-6)] } \\
\geqslant & (x+y+z+12)^{2} \\
\Leftrightarrow & \frac{(x+2)^{2}}{y+z-2}+\frac{(y+4)^{2}}{x+z-4}+\frac{(z+6)^{2}}{x+y-6} \\
\geqslant & \frac{1}{2} \cdot \frac{(x+y+z+12)^{2}}{x+y+z-6}
\end{aligned}$$

Combining with the given equation, we get
$$\frac{(x+y+z+12)^{2}}{x+y+z-6} \leqslant 72$$

When $\frac{x+2}{y+z-2}=\frac{y+4}{x+z-4}=\frac{z+6}{x+y-6}=\lambda$, i.e.,
$$\left\{\begin{array}{l}
\lambda(y+z)-x=2(\lambda+1) \\
\lambda(x+z)-y=4(\lambda+1) \\
\lambda(x+y)-z=6(\lambda+1)
\end{array}\right.$$

the equality in (35) holds.
Let $w=x+y+z+12$. Then
$$\frac{(x+y+z+12)^{2}}{x+y+z-6}=\frac{w^{2}}{w-18}$$

Also,
$$\begin{array}{l} 
\frac{w^{2}}{w-18} \geqslant 4 \times 18=72 \\
\Leftrightarrow w^{2}-4 \times 18 w+4 \times 18^{2} \geqslant 0 \\
\Leftrightarrow(w-36)^{2} \geqslant 0 \\
\frac{(x+y+z+12)^{2}}{x+y+z-6} \geqslant 72
\end{array}$$

When
$$\begin{array}{l}
w=x+y+z+12=36 \\
\Leftrightarrow x+y+z=24
\end{array}$$

the equality in (37) holds.
From (35) and (37), we get
$$\frac{(x+y+z+12)^{2}}{x+y+z-6}=72$$

From the system of equations (36) and (38), we get
$$\left\{\begin{array}{l}
(2 \lambda-1)(x+y+z)=12(\lambda+1), \\
x+y+z=24
\end{array} \Rightarrow \lambda=1\right.$$

Substituting $\lambda=1$ into the system of equations (36), we get
$$\left\{\begin{array}{l}
y+z-x=4 \\
x+z-y=8, \\
x+y-z=12
\end{array} \Rightarrow(x, y, z)=(10,8,6)\right.$$

Therefore, the unique real solution is
$$(x, y, z)=(10,8,6)$$