Finding all possible values of $z$ given three conditions. Suppose that $x,y,z$ are positive integers satisfying $x \le y \le z$, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of $z$?
I have only found $2$ sets of positive integers solutions satisfying all conditions, which are $(x, y, z)=(1, 3, 8), (1, 4, 5).$ Would this mean our final solution is $8+5=13?$ I'm not sure if there are any more pairs, or if I did something wrong.

$\frac{1}{2} = \frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} \leq \frac{3}{x^2}$.
Therefore, $x^2 \leq 6$ which only leaves $x=1$ and $x=2$
For $x=1$, you get $(x,y,z) = (1,3,8)$ and $(1,4,5)$. I'll leave it to you to prove that these are the only possibilities.
Hint: Use the same technique as above to get an upper bound on $y$ and then try all possible values.
Hint2: $yz = 2 (1+y+z) \Rightarrow \frac{1}{2} = \frac{1}{yz} + \frac{1}{z} + \frac{1}{y} \leq \frac{3}{y}$
For $x=2$, you get $(x,y,z) = (2,2,4)$. I'll let you prove that this is the only possibility
So the sum of all $z$ would be 17.
