$4 \cdot 55$  Try to find real numbers $x, y, z$ greater than 1, satisfying the equation
$$
x+y+z+\frac{3}{x-1}+\frac{3}{y-1}+\frac{3}{z-1}=2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}) .
$$
[Solution] 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 $t-1-\sqrt{t+2}=0$.
That is, when $t=\frac{3+\sqrt{13}}{2}$, the equality holds.
It is easy to see that the original equation can be transformed into
$$
f(x)+f(y)+f(z)=0 \text {, where } x>1, y>1, z>1 \text {, which is equivalent to }
$$

$$
f(x)=0, f(y)=0, f(z)=0 .
$$

Therefore, $x=y=z=\frac{3+\sqrt{13}}{2}$.