Let $a,b,c$ be the length of sides of a triangle then prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$ 
Let $a,b,c$ be the length of sides of a triangle then prove that:
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$

Please help me!!!

$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)
=\dfrac{1}{2}[(a+b-c)(b+c-a)(a-b)^2+(b+c-a)(a+c-b)(b-c)^2+(a+c-b)(a+b-c)(c-a)^2]
≥0$$
