Firstly, by AM-GM $$\frac{a+b+c}{3}\ge\sqrt[3]{abc}\implies \frac{1}{3}\ge\sqrt[3]{abc}\implies \frac{1}{abc}\ge 27$$ Therefore, $$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} =\frac{a+b+c}{abc}=\frac{1}{abc}\ge 27$$ Meanwhile, by AM-HM $$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le\frac{a+b+c}{3}=\frac{1}{3}\implies \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 9$$ It follows that, \begin{align*} &\Big(1+\frac{1}{a}\Big)\Big(1+\frac{1}{b}\Big)\Big(1+\frac{1}{c}\Big)\\ =\ & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\\ \ge\ & 1 +9+27+27 \\ =\ & 64 \end{align*}