Prove $a^2b(a-b)+ b^2c(b-c)+ c^2a(c-a)\geq 0$

Question

Given the sides of a triangle a, b, c, prove that $$a^2b(a-b)+ b^2c(b-c)+ c^2a(c-a)\geq 0$$

I can't seem to figure out how to simplify it. I assumed that $a$ is greatest out of the three, but I can't seem to make any headway.