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...
 A: 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. 
A: 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 $x=y=z=2$.
A: [ 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 $x=y=z=2\,$).
A: I will use the cubic equation with roots $x$, $y$, $z$, inspired by the answer of @dxiv:. 
$$X^3 - s X^2 + 12 X - (2+s) =0$$
The discriminant of the equation is $-(2s+13)(2s+15)(s-6)^2$. If $s\ge 0$ then the discriminant is positive only for $s=6$, which gives $a=b=c=2$. Therefore: the system has only one positive solution, but infinitely many negative ones, for each $s \in [-\frac{15}{2}, -\frac{13}{2}]$ we get a solution with $a+b+c = s$. For instance, for $s=-\frac{13}{2}$, we get $a=b=-3$, $c=-\frac{1}{2}$ a solution. 
A: 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 $x=y=z$.
A: 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:
$$x=y=z=2.$$
