Solution:
By the AM-GM inequality we have $a^{3}+b c \geq 2 \sqrt{a^{3} b c} = 2 \sqrt{a^{2}(a b c)} = 2 a$ and
$$\frac{1}{a^{3}+b c} \leq \frac{1}{2 a}$$
Similarly, $\frac{1}{b^{3}+c a} \leq \frac{1}{2 b}$, $\frac{1}{c^{3}+a b} \leq \frac{1}{2 c}$, and then
$$\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{1}{2 a}+\frac{1}{2 b}+\frac{1}{2 c} = \frac{1}{2} \frac{a b+b c+c a}{a b c} \leq \frac{(a b+b c+c a)^{2}}{6}$$
Therefore, it is enough to prove $\frac{(a b+b c+c a)^{2}}{6} \leq \frac{(a b+b c+c a)^{2}}{6}$. This inequality is trivially shown to be equivalent to $3 \leq a b+b c+c a$, which is true because of the AM-GM inequality: $3 = \sqrt[3]{(a b c)^{2}} \leq a b+b c+c a$.