We may homogenise in two steps the inequality we wish to show. First, we use $(a+b+c)^2 = 9$ to replace the RHS and obtain the equivalent inequality
$$a^2 + b^2 + c^2 + 6abc \le (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) + 3abc.$$
This is equivalent to $3abc \le ab + bc + ca$. We now multiply by $3 = a+b+c$ to obtain the equivalent inequality
$$9abc \le (a + b + c)(ab + bc + ca) = a^2b + ab^2 + b^2c + bc^2 + c^2a + a^2c + 3abc,$$
i.e.,
$$6abc \le a^2b + ab^2 + b^2c + bc^2 + c^2a + a^2c, \text{ or}$$
$$6 \le \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{a} + \frac{a}{c},$$
which immediately follows from the AM-GM inequality.