<s>[INST] Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$ 
Solve the following system of equations in $\Bbb R^+$:
   $$
\left\{ 
\begin{array}{l}
xy+yz+xz=12 \\ 
xyz=2+x+y+z\\ 
\end{array}
\right. 
$$  

I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
 [/INST] I will show that the only positive solution is indeed $(2,2,2).$ $$\dfrac{xyz}{xy+yz+zx} = \dfrac{1}{6}+\dfrac{x+y+z}{12} = \dfrac{\sqrt{xy+yz+zx}}{12\sqrt{3}}+\dfrac{x+y+z}{12}\geq\dfrac{xyz}{4(xy+yz+zx)}+\dfrac{3xyz}{4(xy+yz+zx)} = \dfrac{xyz}{xy+yz+zx}$$, by AM-GM inequalities provided that all of them are non-negative. 
Therefore, either the problem intended to ask for only non-negative solutions, or it has infinitely many solutions indicated by the answer above. 
</s>