By the AM-GM inequality, we have
$$a^3 + b^2 + b \ge 3\sqrt[3]{a^3b^3} = 3ab,$$
$$b^2 + b + 1 \ge 3\sqrt[3]{b^3} = 3b.$$
This gives $a^3 + 2b^2 + 2b + 4 \ge 3ab + 3b + 3 = 3(ab + b + 1)$. Since $abc = 1$, we can let $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$ for some positive real numbers $x, y, z$. Then we have
$$\frac{1}{a^3 + 2b^2 + 2b + 4} \le \frac{1}{3(ab + b + 1)} = \frac{z}{3(x + y + z)}. \quad (1)$$
By symmetry, we have
$$\frac{1}{b^3 + 2c^2 + 2c + 4} \le \frac{x}{3(y + z + x)}, \quad (2)$$
$$\frac{1}{c^3 + 2a^2 + 2a + 4} \le \frac{y}{3(z + x + y)}. \quad (3)$$
Adding (1), (2) and (3), we obtain the desired inequality. Equality holds when $a = b = c = 1$.