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?

Question

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 $

Answer 1

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.