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...

On one hand, we have $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$, then one has $$-8 \leq x+ y +z \leq 6.$$
On other hand, we have $$(x+y+z)^2 \geq 3(xy+yz+zx) = 36,$$
then $$x+y+z \geq 6$$ or $$x+y+z \leq -6.$$
So, you have $x+y+z = 6$ or $-8 \leq x+y+z \leq -6$ (but x, y, z is non-negative!).
Thus, $x+y+z = 6$ and $x=y=z$.
