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!!!

Label the triangle so that $a>b>c>0$
Then $ab > b^2$ and $cb> c^2$ and $ac> a^2$
So $ab-b^2 > 0$ and $a^2(ab-b^2)>0$
Similarly for the other two terms, so that their sum will naturally be greater than $0$
