From $\frac{x}{y} > \frac{a}{b}$ follows that $xb > ya$, hence $xb - ya \ge 1$. In the same way, $at - bz \ge 1$. Multiplying the first inequality by $t$, the second one by $y$ and adding the two relations yields $bxt - byz \ge t + y$, that is $b \ge t + y$.

In the same way, $a \ge x + z$. These two inequalities lead to the conclusion.

If $d = \text{g.c.d.}(a, b)$, then $a = d a_1$ and $b = d b_1$, with $a_1, b_1 \in \mathbb{N}^*$, $(a_1, b_1) = 1$ and $\frac{x}{y} > \frac{a_1}{b_1}$. It follows that $b_1 x - a_1 y = u \ge 1$ and $a_1 t - b_1 z = v \ge 1$, with $u, v \in \mathbb{N}$. The first relation gives $b_1 x z - a_1 y z = u z$, and the second gives $a_1 x t - b_1 z x = v x$. Adding these relations yields $a_1 = u z + v x \ge z + x$. In the same way, $b_1 = u t + v y \ge t + y$. Since $ab \ge a_1 b_1$, the conclusion is proven.