Subtract $9$ from both sides and use the well-known identity
$$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$

$$\frac{a^3 + b^3 + c^3}{abc} - 3 \ge 9 \cdot \frac{a^2 + b^2 + c^2}{ab + bc + ca} - 9 \Leftrightarrow \\ \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc}{abc} - 3 \ge 9 \cdot \frac{a^2+b^2+c^2}{ab+bc+ca} - 9 \Leftrightarrow$$

$$\begin{aligned} \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc-3abc}{abc} &\ge \frac{9(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca} \\ &\iff \\ \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{abc} &\ge \frac{9(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca} \\ &\iff \\ \frac{(a^2+b^2+c^2-ab-bc-ca)((a+b+c)(ab+bc+ca)-9abc)}{abc(ab+bc+ca)} &\ge 0. \end{aligned}$$

The last inequality holds, since
$$a^2 + b^2 + c^2 - ab - bc - ca \ge 0 \quad \text{and} \quad (a+b+c)(ab+bc+ca) - 9abc \ge 0.$$
By the inequality with the sum of three squares:
$$\begin{aligned} (a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0 &\iff a^2 + b^2 + c^2 \ge ab + bc + ca \\ (a+b+c)^2 \ge 3(ab+bc+ca) &\Rightarrow (a+b+c)^6 \ge 3^3(ab+bc+ca)^3 \ge 27 \cdot 27(abc)^2 \\ (a+b+c)^3 &\ge 27abc. \end{aligned}$$
$$\text{Moreover, } \left(\frac{1}{3}(x+y+z)\right)^3 \ge xyz \Rightarrow (ab+bc+ca)^3 \ge 27a^2b^2c^2.$$
$$\text{Hence } (a+b+c)^3(ab+bc+ca)^3 \ge (9abc)^3.$$