<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] The condition gives $$xyz=2+x+y+z$$ or
$$xyz=2+\frac{(x+y+z)(xy+xz+yz)}{12}$$ or
$$12xyz=24+\sum_{cyc}(x^2y+x^2z+xyz)$$ or
$$3xyz=24+\sum_{cyc}(x^2y+x^2z-2xyz),$$
which gives
$$3xyz-24=\sum_{cyc}(x^2y+x^2z-2xyz)=\sum_{cyc}z(x-y)^2\geq0.$$
Thus, $xyz\geq8$ and the equality occurs for $x=y=z$.
On the other hand, by AM-GM
$$12=xy+xz+yz\geq3\sqrt[3]{x^2y^2z^2},$$
which gives $xyz\leq8.$
Thus, $xyz=8$ and we get $x=y=z=2$.
</s>