First, we will prove if $\forall x, y > 0$ then the inequality $(*) \frac{x^2-xy+y^2}{x^2+xy+y^2} \geq \frac{1}{3}$ holds.

$(*) \Leftrightarrow 3(x^2-xy+y^2) \geq x^2+xy+y^2 \Leftrightarrow 2(x^2+y^2)-4xy \geq 0 \Leftrightarrow 2(x-y)^2 \geq 0$ thus proof of $(*)$ completes. Consequently,
$$\frac{a^2(a^3 + b^3)}{a^2 + ab + b^2} = a^2(a + b) \frac{a^2 - ab + b^2}{a^2 + ab + b^2} \geq \frac{1}{3} a^2(a + b) \text{ by }(*).$$
Similarly, $\frac{b^2(b^3+c^3)}{b^2+bc+c^2} \geq \frac{1}{3}b^2(b+c)$ and $\frac{c^2(c^3+a^3)}{c^2+ca+a^2} \geq \frac{1}{3}c^2(c+a)$ from where follows
$$\frac{a^2(a^3+b^3)}{a^2+ab+b^2} + \frac{b^2(b^3+c^3)}{b^2+bc+c^2} + \frac{c^2(c^3+a^3)}{c^2+ca+a^2} \geq \frac{1}{3}(a^3+b^3+c^3) + \frac{1}{3}(a^2b+b^2c+c^2a) \geq \frac{1}{3} \cdot 3\sqrt{a^3b^3c^3} + \frac{1}{3} \cdot 3\sqrt{a^2b \cdot b^2c \cdot c^2a} = 2abc \text{ and we have done. Equality holds when } a=b=c.$$