After expansion it's just $2(a^3+b^3+c^3)\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$ which is obvious.

Prove that for any positive real numbers $a, b, c$, $$a^3 + b^3 + c^3 + 3abc \ge \frac{3}{4}(a+b)(b + c)(c + a)$$$$a^3 + b^3 + c^3 +6abc \ge \frac{9}{8}(a+b)(b + c)(c + a)$$