, 2014

By setting $a = b = c$, it follows that $k \leq \sqrt[3]{9} - 1$. We claim that $k = \sqrt[3]{9} - 1$ is the largest possible number so that the inequality holds. Let
$$A = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}, \quad B = \frac{b}{a} + \frac{c}{b} + \frac{a}{c}.$$
By AM-GM inequality, we get that
$$A \geq 3\sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3,$$
and similarly,
$$B \geq 3\sqrt[3]{\frac{b}{a} \cdot \frac{c}{b} \cdot \frac{a}{c}} = 3.$$
For each real number $k \geq 0$, the following inequalities are true:
$$9(k^3 + 1) \leq (k^3 + 1)AB$$
$$9k^2A \leq 3k^2AB$$
$$9kB \leq 3kAB.$$

Adding these inequalities we get
$$9(k^3 + 1 + k^2 A + k B) \le (k + 1)^3 AB \\ 9 \left(k + \frac{a}{b}\right) \left(k + \frac{b}{c}\right) \left(k + \frac{c}{a}\right) \le (k + 1)^3 AB.$$
Substituting $k = \sqrt[3]{9} - 1$ in the last inequality, we discover that
$$9 \left(k + \frac{a}{b}\right) \left(k + \frac{b}{c}\right) \left(k + \frac{c}{a}\right) \le 9AB.$$
which is precisely, the desired inequality. Therefore the largest possible $k$ is $\sqrt[3]{9} - 1$ as claimed. $\square$