Let $a,$ $b,$ $c$ be positive real numbers.  Find the smallest possible value of
\[6a^3 + 9b^3 + 32c^3 + \frac{1}{4abc}.\]
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 $\boxed{6}.$