$x+y≤xy$ and $y+z≤yz$ implies $x+z≤xz$

Question

Let $x,y$ and z $\in \mathbb{Z}\setminus \{1\}$ such that $x+y\leqslant xy$ and $y+z\leqslant yz$. Prove that $x+z\leqslant xz$

Answer 1

Recall that $(a - 1)(b - 1) = ab - a - b + 1$, so $x + y \le xy \iff (x - 1)(y - 1) \ge 1$. Thus, we are assuming \begin{align*} 1 &\le (x - 1)(y - 1) \\ 1 &\le (y - 1)(1 - 1), \end{align*} and trying to prove $$1 \le (x - 1)(z - 1).$$ Since these are non-zero integers, being $\ge 1$ is equivalent to being $> 0$. So, our assumptions are therefore that $x - 1$ and $y - 1$ have the same sign, as do $y - 1$ and $z - 1$. We therefore conclude that $x - 1$ and $z - 1$ have the same sign too, proving what we needed.