jmc

By AM-GM, $$6a^3+9b^3+32c^3\ge 3\sqrt[3]{6a^3\cdot 9b^3\cdot 32c^3}=36abc.$$Again by AM-GM, $$36abc+\frac{1}{4abc}\ge 2\sqrt{36abc\cdot \frac{1}{4abc}}=6.$$Equality occurs when $6a^3=9b^3=32c^3$ and $36abc=3.$ We can solve, to get $a=\frac{1}{\sqrt[3]{6}},$ $b=\frac{1}{\sqrt[3]{9}},$ and $c=\frac{1}{\sqrt[3]{32}}.$ Therefore, the minimum value is $\boxed{6}.$