<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] Using AM-GM:
$$12=xy+yz+xz\ge 3\sqrt[3]{(xyz)^2} \Rightarrow xyz\le 8 \ \ \ \ (1)$$
$$xyz=2+x+y+z\le8 \Rightarrow x+y+z\le 6 \ \ \  \ \ (2)$$
The equality occurs when $x=y=z=2$.
Using C-S:
$$(x^2+y^2+z^2)(y^2+z^2+x^2)\ge (xy+yz+zx)^2 \Rightarrow x^2+y^2+z^2\ge 12 \ \ \ \ (3)$$
The equality occurs when $x=y=z=2$.
From the first equation:
$$(x+y+z)^2=24+x^2+y^2+z^2\ge 36 \Rightarrow x+y+z\ge 6 \ \ \ \ \ (4)$$
From $(2)$ and $(4)$, and then from the second equation we get:
$$x+y+z=6, \\ xyz=8.$$
So, we get:
$$x=y=z=2.$$
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