Blackboard operation $x,y,z\rightarrow x,y,1/(zx+zy)$ The three numbers $2,3,6$ are written on the blackboard. In each move, we can pick any two numbers, say $x,y$, and replace the third number $z$ by $1/(zx+zy)$. Using finitely many operations, is it possible to obtain the three numbers $2,3,4$?
In the first move, we can get $2,3,1/30$, and if we fix $2,3$ again, we get back $2,3,6$. To prove that it's impossible to reach $2,3,4$, we might need to define some potential function. But it's not clear which.
[Source: Russian competition problem]

The invariance is motivated by $ \frac{ x+y} { z ( x + y) } = \frac{ 1}{z} $. 
This suggests that we want $ xz, yz, \frac{1}{z}$ as terms in the invariance. Of course, it needs to be cyclic, so the potential invariance has the form
$$ A ( xy+yz+zx) + B ( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) $$
Hint: The invariance is

 $$ f(x,y,z) =  ( xy+yz+zx) + ( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) $$

You can verify this because

 $$ f( x,y,\frac{1}{ z (x+y) }) = ( xy + \frac{1}{z} ) + ( \frac{1}{x} + \frac{1}{y} + zx+zzy) $$

Now, verify that $ f(2, 3, 4) \neq f (2,3, 6 ) $.

