Solution:
Let $x$ be a positive real number. By AM-GM we have $\frac{1+x+x+x}{4} \geq x^{\frac{3}{4}}$, or equivalently $1+3 x \geq 4 x^{\frac{3}{4}}$. Using this inequality we obtain:
$$\left(3 a^{2}+1\right)^{2} \geq 16 a^{3} \text{ and } 2\left(1+\frac{3}{b}\right)^{2} \geq 32 b^{-\frac{3}{2}}$$
Moreover, by inequality of arithmetic and geometric means we have
$$f(a, b)=\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2} \geq 16 a^{3}+32 b^{-\frac{3}{2}}=16\left(a^{3}+b^{-\frac{3}{2}}+b^{-\frac{3}{2}}\right) \geq 48 \frac{a}{b}$$
Therefore, we obtain
$$f(a, b) f(b, c) f(c, a) \geq 48 \cdot \frac{a}{b} \cdot 48 \cdot \frac{b}{c} \cdot 48 \cdot \frac{c}{a}=48^{3}$$
Equality holds only when $a=b=c=1$.