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

$$x^{2}+y^{2} \geq 2 x y+1, \quad y^{2}+z^{2} \geq 2 y z+4, \quad x^{2}+z^{2} \geq 2 x z+9$$

or otherwise

$$z x^{2}+z y^{2} \geq 2 x y z+z, \quad x y^{2}+x z^{2} \geq 2 x y z+4 x, \quad y x^{2}+y z^{2} \geq 2 x y z+9 y$$

Adding up the last three inequalities we have

$$x y(x+y)+y z(y+z)+z x(z+x) \geq 6 x y z+4 x+9 y+z$$

which implies that $(x+y+z)(x y+y z+z x-2) \geq 9 x y z+2 x+7 y-z$.

Since $x \geq z+3$ it follows that $2 x+7 y-z \geq 0$ and our inequality follows.

Case 2. $y=z+1$. Since $x \geq y+1=z+2$ it follows that $x \geq z+2$, and replacing $y=z+1$ in the required inequality we have to prove

$$(x+z+1+z)(x(z+1)+(z+1) z+z x-2) \geq 9 x(z+1) z$$

which is equivalent to

$$(x+2 z+1)\left(z^{2}+2 z x+z+x-2\right)-9 x(z+1) z \geq 0$$

Doing easy algebraic manipulations, this is equivalent to prove

$$(x-z-2)(x-z+1)(2 z+1) \geq 0$$

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 $k$.

## Combinatorics

The text is already in English, so no translation is needed.