Example 13 (1992 Nordic Competition Problem) Find real numbers $x, y, z$ greater than 1 that satisfy the equation
$$
x+y+z+\frac{3}{x-1}+\frac{3}{y-1}+\frac{3}{z-1}=2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}) .
$$
Let $f(t)=t+\frac{3}{t-1}-2 \sqrt{t+2}$, then
$$
\begin{aligned}
f(t) & =\frac{1}{t-1}\left[t^{2}-t+3-2(t-1) \sqrt{t+2}\right] \\
& =\frac{1}{t-1}\left[t^{2}-2 t+1+(\sqrt{t+2})^{2}-2(t-1) \sqrt{t+2}\right] \\
& =\frac{1}{t-1}[(t-1)-\sqrt{t+2}]^{2} .
\end{aligned}
$$

Obviously, when $t>1$, $f(t) \geqslant 0$, and the equality holds only when $t-1-\sqrt{t+2}=0$, i.e., when $t=\frac{1}{2}(3+\sqrt{13})$.

It is easy to see that the original equation can be transformed into $f(x)+f(y)+f(z)=0$, where $x>1, y>1, z>1$, which is equivalent to $f(x)=0, f(y)=0, f(z)=0$, so $x=y=z=\frac{1}{2}(3+\sqrt{13})$ is the solution.