By the conditions, $\tfrac1{ab}+\tfrac1{bc}+\tfrac1{ca}=1$. Assume each of $a$, $b$, $c$ is less than or equal to $\tfrac{17}{10}$, then \[\frac1{ab}+\frac1{bc}+\frac1{ca}\ge3\times\frac1{(17/10)^2}=\frac{300}{289}>1,\]which contradicts. Therefore, at least one of $a, b$ or $c$ is greater than $\tfrac{17}{10}$ .