The inequality is equivalent to
$$8(a+b+c)(ab+bc+ca) \leq 9(a+b)(b+c)(c+a)$$
Using the identity
$$(a+b+c)(ab+bc+ca) = (a+b)(b+c)(c+a) + abc$$
we get
$$8(a+b)(b+c)(c+a) + 8abc \leq 9(a+b)(b+c)(c+a),$$
hence $8abc \leq (a+b)(b+c)(c+a)$. Applying AM-GM inequality, it follows
$$2\sqrt{ab} \leq a+b, \quad 2\sqrt{bc} \leq b+c, \quad 2\sqrt{ca} \leq c+a,$$
and by multiplication $8abc \leq (a+b)(b+c)(c+a)$. The equality holds if and only if $a = b = c$.