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...

[ EDIT ] The question has been changed to require $\,x,y,z \ge 0\,$ now, which renders the following not applicable any longer.

(Too long for a comment.) Let $x=a+2, y=b+2, z=c+2\,$, then the conditions rewrite as:

$$\require{cancel}
\begin{cases}
\begin{align}
a b + b c + a c + 4(a + b + c) + \cancel{12} &= \cancel{12} \\
a b c + 2(a b + a c + b c) + 4(a + b + c) + \cancel{8} &= a+b+c + \cancel{8}
\end{align}
\end{cases}
$$
$$\require{cancel}
\iff\;\; \begin{cases}
\begin{align}
a b + b c + a c + 4(a + b + c) &= 0 \\
a b c + 2(a b + a c + b c) + 3(a + b + c) &= 0
\end{align}
\end{cases}
$$
Let $u = a+b+c\,$, then:
$$\require{cancel}
\begin{align}
a b + b c + a c &= -4 u \\
a b c  &= 5u\end{align}
$$
Therefore $a,b,c$ are the roots of $t^3-ut^2-4 ut-5u\,$, and the problem reduces to finding for what values of $u$ the cubic has three real roots (other than $u=0$ which corresponds to $x=y=z=2\,$).
