Since $x$, $y$, $z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:

Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$(x-y)^2 \geq 1, \quad (y-z)^2 \geq 4, \quad (x-z)^2 \geq 9$$
which are equivalent to
$$x^2+y^2 \geq 2xy+1, \quad y^2+z^2 \geq 2yz+4, \quad x^2+z^2 \geq 2xz+9$$
or otherwise
$$\geq x^2 - 2y^2 \geq 2xyz + z, \quad xy^2 + xz^2 \geq 2xyz + 4x, \quad yx^2 + yz^2 \geq 2xyz + 9y$$
Adding up the last three inequalities we have
$$xy(x+y)+yz(y+z)+zx(z+x) \geq 6xyz+4x+9y+z$$
which implies that $(x+y+z)(xy+jz+zx-2) \geq 9xyz+2x+7y-z$.
Since $x \geq z+3$ it follows that $2x+7y-z \geq 0$ and our inequality follows.

Case 2. $y=z+1$. Since $x \geq y+1 \geq z+2$ it follows that $x \geq z+2$, and replacing $y=z+1$ in the required inequality we have to prove
$$(x+z+1+z)(x(z+1)+(z+1)z+xz-2) \geq 9x(z+1)z$$
which is equivalent to
$$(x+2z+1)(z^2+2zx+z+x-2)-9x(z+1)z \geq 0$$
Doing easy algebraic manipulations, this is equivalent to prove
$$(x-z-2)(x-z+1)(2z+1) \geq 0$$
which is satisfied since $x \geq z+2$.
The equality is achieved only in the Case 2 for $x=z+2$, so we have equality when $(x,y,z)=(k+2,k+1,k)$ and all the permutations for any positive integer $k$.