
92.1. Determine all real numbers $x>1, y>1$, and $z>1$, satisfying the equation

$$
\begin{aligned}
x+y+z+\frac{3}{x-1} & +\frac{3}{y-1}+\frac{3}{z-1} \\
& =2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2})
\end{aligned}
$$


Solution. Consider the function $f$,

$$
f(t)=t+\frac{3}{t-1}-2 \sqrt{t+2}
$$

defined for $t>1$. The equation of the problem can be written as

$$
f(x)+f(y)+f(z)=0
$$

We reformulate the formula for $f$ :

$$
\begin{aligned}
f(t) & =\frac{1}{t-1}\left(t^{2}-t+3-2(t-1) \sqrt{t+2}\right) \\
& =\frac{1}{t-1}\left(t^{2}-2 t+1+(\sqrt{t+2})^{2}-2(t-1) \sqrt{t+2}\right) \\
& =\frac{1}{t-1}(t-1-\sqrt{t+2})^{2}
\end{aligned}
$$

So $f(t) \geq 0$, and $f(t)=0$ for $t>1$ only if

$$
t-1=\sqrt{t+2}
$$

or

$$
t^{2}-3 t-1=0
$$

The only $t$ satisfying this condition is

$$
t=\frac{3+\sqrt{13}}{2}
$$

So the only solution to the equation in the problem is given by

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