Tricky proof of inequality

Question

Positive numbers a, b and c are given. How one can prove that $$a^3 + b^3 + c^3 + ab^2 + bc^2 + ca^2\geq2(a^2b + b^2c + c^2a)$$

Answer 1

Try to prove that $a^3 + ab^2 \ge 2a^2b$, and similarly $b^3 + bc^2 \ge 2b^2c$ and $c^3+ca^2 \ge 2c^2a$.

Answer 2

$$ a^3 + b^3 + c^3 + ab^2 + bc^2 + ca^2\geq2(a^2b + b^2c + c^2a) $$ is equivalent to $$ a*(a^2 + b^2-2ab) + b*(b^2+c^2-2bc) + c*(c^2+a^2-2ac)\geq 0 $$ or $$ a*(a-b)^2 + b*(b-c)^2 + c*(c-a)^2\geq 0 $$ which is true.