But They're Integers!

Assuming that none of $x,y,z=0$ because we are trying to find the maximum value.On first sight I thought that $xyz+xy+yz+zx$ has a relation to $(x+1)(y+1)(z+1)$ and it did.From $A.M\geq G.M$ we have $:\\\dfrac{(x+1)+(y+1)+(z+1)}{3}\geq [\sqrt[3]{(x+1)(y+1)(z+1)}]$ ,this simplifies to $(\dfrac{14}{3})^{3}\geq (xyz+xy+yz+zx+11+1)\\ \Rightarrow (\dfrac{14}{3})^{3}-12\geq(xyz+xy+yz+zx).$

So, the maximum possible value with real $x,y,z$ is 89.62, where $x=y=z$ , but given that $x,y,z$ are integers, we will plug the closest possible $x,y,z$ , which is $3,4,4$ to get the maximum as $\boxed{88}$ .

Perhaps, this is more straightforward.

$(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)$

$\Rightarrow xyz + xy+yz+zx = xyz + \dfrac {(x+y+z)^2 - (x^2+y^2+z^2)} {2}$

$\quad = xyz + \dfrac {121}{2} - \dfrac { x^2+y^2+z^2} {2} \le xyz + \frac {121}{2} - \frac {3}{2} \left( \sqrt [3] {xyz} \right)^2$

From $\dfrac {x+y+z} {3} = \frac {11}{3} \ge \sqrt [3] {xyz}$ , therefore,

$\Rightarrow xyz + xy+yz+zx \le \left( \frac {11}{3} \right)^3 + \frac {121}{2} - \frac {3}{2} \left( \frac {11}{3} \right)^2 = 89.62962963$

And the nearest integer solution is $\boxed{88}$ .

Let z = 11 - x - y. Then xyz + xy + yz + zx = xy(11 - x - y) + xy + y(11 - x - y) + x(11 - x - y). Defining this as f(x,y), df/dx = 11y - 2xy - y^2 + 11 - 2x - y = 0 and equations, 12(y - x) = (y - x)(y + x). Either x = y, or x + y = 12. If x + y = 12, z = -1, which violates the problem conditions, so x = y. Consider the following table: x y z xyz xy yz zx sum --- --- ---- -------- ----- ----- ------ ------- 1 1 9 9 1 9 9 28 2 2 7 28 4 14 14 60 3 3 5 45 9 15 15 84 4 4 3 48 16 12 12 88 5 5 1 25 25 5 5 60 Ed Gray

i answered by instinct (4,3,4)