Putting these equal to 0, we have

..... (1)

..... (2)

..... (3)

Subtract (2) from (1): . Therefore either or . Similarly

either or ; and

either or .

From those equations it clearly follows that x, y and z cannot all be different. On the other hand, they cannot all be the same, because then the equation x+y+z=3 would mean that they are all equal to 1, which would contradict the equation yz+zx+xy=1.

Thus two of x,y and z are equal, and the third is different, say z=y≠x. Then the equations yz+zx+xy=1 and x+y+z=3 become , . Substitute the value of x from the second of those equations into the first to get a quadratic for y. The numbers don't come out particularly pleasantly, but I get the maximum value of V as .