Find all real numbers $x, y, z$ satisfying:

$$
\left\{\begin{array}{l}
(x+1) y z=12 \\
(y+1) z x=4 \\
(z+1) x y=4
\end{array}\right.
$$
Let $(x, y, z)$ be a solution. Clearly, none of these numbers is zero. By subtracting the third equation from the second equation, we deduce that $z x=x y$, and by simplifying by $x$ (which is non-zero), we obtain that $z=y$. By subtracting the third equation from the first equation, we get: $y^{2}-x y=8$, or equivalently $x y=y^{2}-8$. The second equation can be rewritten as $y^{2} x+x y=4$. Therefore, we have:

$$
y\left(y^{2}-8\right)+y^{2}-8=4
$$

or equivalently $y^{3}+y^{2}-8 y-12=0$. We notice that $y=3$ is a solution. By performing the Euclidean division of $y^{3}+y^{2}-8 y+12$ by $y-3$, we find:

$$
y^{3}+y^{2}-8 y-12=(y-3)\left(y^{2}+4 y+4\right)=(y-3)(y+2)^{2}
$$

We deduce that $y=z=3$ or $y=z=-2$. In the first case, $x=\frac{1}{3}$ and in the second case, $x=2$. Conversely, the triplets $(2,-2,-2)$ and $\left(\frac{1}{3}, 3,3\right)$ are solutions and these are the only ones.