an inequality over integers Let $x$, $y$ and $z$ be integers different from 1.
Prove that if $x+y\leq xy$ and $y+z \leq yz$ then $x+z\leq xz$
My attempt:
$x+2y+z\leq (x+z)y$ $~~$so $2y \leq (x+z)(y-1)$
and I'm stack here.
I need just some hint.
Thanks.

We have
$$\left\{\begin{aligned} & x+y \leqslant xy \\& y+z \leqslant yz \end{aligned}\right. \Rightarrow \left\{\begin{aligned} & (x-1)(y-1) \geqslant 1\\& (y-1)(z-1) \geqslant 1\end{aligned}\right. \Rightarrow (x-1)(z-1)(y-1)^2 \geqslant 1 > 0.$$
Therefore
$$ (x-1)(z-1) > 0 $$
But $x,\,y,\,z$ are integers, so
$$(x-1)(z-1) \geqslant 1 \Rightarrow x+z \leqslant xz.$$
Done.
