Prove that, $a^3+b^3+c^3+3abc>a^2b+ab^2+b^2c+bc^2+a^2c+ac^2$

Question

If $a,b,c$ are triangle sides, Prove that,

$$a^3+b^3+c^3+3abc≥a^2b+ab^2+b^2c+bc^2+a^2c+ac^2$$

I started from $a^3+b^3+c^3≥3abc$, But and I saw that this method did not work.

Answer 1

$$a^3+b^3+c^3+3abc-a^2b-ab^2-b^2c-bc^2-a^2c-ac^2≥0 \Leftrightarrow \frac 12×\left((a-b)^2(a+b-c)+(a-c)^2(a+c-b)+(b-c)^2(b+c-a) \right)≥0$$