(IMO Shortlist 2004 A5 modified) By the power mean inequality, we have
$$\begin{aligned} S &:= \sqrt[4]{\frac{\sqrt{3}}{a} + 6\sqrt{3}b} + \sqrt[4]{\frac{\sqrt{3}}{b} + 6\sqrt{3}c} + \sqrt[4]{\frac{\sqrt{3}}{c} + 6\sqrt{3}a} \\ &\le 3\sqrt[4]{\frac{1}{3}\left(\frac{\sqrt{3}}{a} + 6\sqrt{3}b + \frac{\sqrt{3}}{b} + 6\sqrt{3}c + \frac{\sqrt{3}}{c} + 6\sqrt{3}a\right)} \\ &= 3^{\frac{7}{8}}\sqrt[4]{\frac{(ab + bc + ca) + 6abc(a + b + c)}{abc}}. \end{aligned}$$
Note that $ab + bc + ca = 1$ and
$$abc(a + b + c) = (ab)(bc) + (bc)(ca) + (ca)(ab) \le \frac{1}{3}(ab + bc + ca)^2 = \frac{1}{3}.$$
Therefore, we have $S \le 3^{\frac{7}{8}}\sqrt[4]{\frac{3}{abc}} = 3^{\frac{9}{8}}(abc)^{-\frac{1}{4}}$. It remains to prove $abc \le 3^{-\frac{3}{2}}$.
Indeed, by the AM-GM inequality, we have
$$1 = ab + bc + ca \ge 3\sqrt[3]{a^2 b^2 c^2}.$$
This clearly implies $abc \le 3^{-\frac{3}{2}}$. This proves the desired inequality. Equality holds when $a = b = c = \frac{1}{\sqrt{3}}$.