, 2016

The minimum value is $\frac{27}{2}$.
Let $x = \frac{1}{a}$, $y = \frac{1}{b}$ and $z = \frac{1}{c}$. Then we have $xyz = 1$. By the AM-GM inequality, we have
$$a^3 + 2 = a^3 + 1 + 1 \ge 3\sqrt[3]{(a^3)(1)(1)} = 3a.$$
Therefore,
$$\frac{a^3 + 8}{a^3(b+c)} \ge \frac{3a + 6}{a^3(b+c)} = \frac{3a^2bc + 6abc}{a^3(b+c)} = \frac{3x + 6x^2}{y+z}.$$
Similarly, we have $\frac{b^3 + 8}{b^3(c+a)} \ge \frac{3y + 6y^2}{z+x}$ and $\frac{c^3 + 8}{c^3(a+b)} \ge \frac{3z + 6z^2}{x+y}$. Thus, it remains to minimize
$$3\left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y}\right) + 6\left(\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}\right).$$
By Nesbitt's inequality, we have
$$\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \ge \frac{3}{2}.$$
Also, by the Cauchy-Schwarz inequality, we have
$$((y+z) + (z+x) + (x+y)) \left( \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \right) \ge (x+y+z)^2.$$
This implies
$$\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{x+y+z}{2} \ge \frac{3\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
by the AM-GM inequality. Therefore, the given expression is at least
$$3 \cdot \frac{3}{2} + 6 \cdot \frac{3}{2} = \frac{27}{2}.$$

Equality holds when $a = b = c = 1$.