For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that? For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that? 
I've managed to find only a lower bound, from $ xyz=2+x+y+z \le \frac{(x+y+z)^3}{27} $, from which I got $x+y+z \ge 6 $
 A: No, there is no upper bound. We solve for $x$ and get
$$ x = \frac{2+y+z}{yz - 1} $$
Pick $z=\frac{2}{y}$, then we get
$$ x = 2 + y + \frac{2}{y} $$
and hence for this choice
$$ x+y+z = 2 + 2y + \frac{4}{y} $$
Choosing $y$ large will get you a large number for $x+y+z$.
Added: To make it more fun. Let us consider the same question when $x,y,z\geq 2$. For that case we get indeed a bound. Wlog we may assume that $2\leq x\leq y \leq z$. Then we get
$$ y = \frac{2+x+ z}{xz -1} \leq \frac{2+2z}{2 z-1} \leq \frac{6}{3} =2 $$
and the way we get $x=2$. Thus, we get
$$ z = \frac{2+x+y}{xy -1} = \frac{6}{3} =2 $$
Thus, $(x,y,z)=(2,2,2)$ is the only solution and hence, $x+y+z=6$.
Let us now consider the case, where we only allow positive integer solutions. By the argument above, we only have to consider the case where $x=1$ otherwise we automatically get $x=y=z=2$. We narrow down the possibilites for $y$. We have
$$ y  = \frac{2+x+z}{xz -1} = \frac{3+z}{z-1} =:f(z)$$
Note that $f$ is decreasing on $(1;\infty)$. We get
$$ f(2)=5, f(3)=3 $$
The first case cannot happen, as we assumed $y\leq z$. Hence, $y\in \{ 1, 2, 3\}$. Now we simply test which combinations solve our equation. 
For $y=1$ we get
$$ z = 4 + z $$
which has no solution.
For $y=2$ we get
$$ 2z = 5 + z $$
hence, we get the valide solution $(1; 2; 5)$.
For $y=3$ we get
$$ 3z = 6 + z $$
which yields $z=3$ and hence another solution, namely $(1; 3; 3)$.
Therefore, we have that - up to permuting the entries - that we only have the solutions
$$ (2;2 ;2), \ (1; 2;5), \ (1;3; 3) $$
Thus, we get that $x+y+z\leq 8$ if we assume that $x,y,z$ are positive integers.
