KVPY SA Exam 2013 Question Paper

Taking three no's.
x + 1, y + 1, z + 1
AM ≥ GM
$\frac{\left(\mathit{x}+1\right)+\left(\mathit{y}+1\right)+\left(\mathit{z}+1\right)}{3}\ge$ ${\left\{\left(\mathit{x}+1\right)\left(\mathit{y}+1\right)\left(\mathit{z}+1\right)\right\}}^{1∕3}$
${\left(\frac{13}{3}\right)}^3$ ≥ xyz + xy + yz + zx + 11
${\left(\frac{13}{3}\right)}^3-11$ ≥ xyz + xy + yz + zx
equality hold when x = y = z but x + y + z = 0 and x, y, z are integers.
So maximum value when any two of x, y, z are equal to 3 and third is equal to 4.