Mathematica competitions in Croatia

Assume on the contrary that $\frac{rs}{t}$, $\frac{rt}{s}$, $\frac{st}{r}$ are positive integers. Without loss of generality we may assume that $a<b<c$. Since $\frac{rs}{t}$ is a positive integer, $\frac{r^2{s}^2}{t^2}$ is also a positive integer, and hence $$\frac{r^2{s}^2}{t^2}=\frac{a^{2}bc+ab+ac+1}{bc+1}=a^2+\frac{ab+ac+1-a^2}{bc+1}$$ is also a positive integer. Since $a<b<c$, we have $$ab+ac+1-a^2>ac+1=s^2>0.$$ Therefore, both the numerator and the denominator of the fraction $\frac{ab+ac+1-a^2}{bc+1}$ are positive, so $$ab+ac+1-a^2\ge bc+1⟹(b-a)(c-a)\le 0,$$ which is a contradiction.