Integers here! Max?

For max. Value all must be around same values ( what I apply to All problems) thus I tried 3,3,4 and there is was. :)

Note that the first condition is equivalent to $(x+1)+(y+1)+(z+1)=13$

To answer the question would be the same as finding the maximum possible value of

$xyz+xy+yz+zx+x+y+z+1$

$=xyz+xy+yx+zx+11$

$=(x+1)(y+1)(z+1)$ , with $(x+1),(y+1),(z+1)$ all being positive integers.

Now assume w.l.o.g. that $x\leq y\leq z$ or equivalently $x+1 \leq y+1 \leq z+1$ .

Checking for the cases $x+1=1,2, \dots, 4$ (since if $x \geq 4$ , we have $x+y+z \geq 12$ , a contradiction), we have the case $x=3, y=3, z=4$ giving a maximum possible value for $xyz+xy+yz+zx$ as $(3+1)(3+1)(4+1)-11=\boxed{69}$