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: let
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=0$      (1)
let
$x=-a+b+c; y=a-b+c; z=a+b-c$
$z,y,x$ are twice the length of the segments between the vertices and the touching point of the incircles. so
$a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}$
substitute them to (1) and multiply the inequality by 16
$(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)+(x+y)^2(y+z)(x-z)\geqslant 0$
$x^3z+y^3x+z^2y\geqslant x^2yz+y^2zx+z^2xy$
and so
$x^3z+y^3x+z^3y-x^2yz-y^2zx-z^2xy$
$=zx(x-y)^2+xy(y-z)^2+yz(z-x)^2\geqslant 0$
as $x>0, y>o, z>0$ equality holds if and only if $x=y=z$.
example if $a=b=c$ then triangle is equilateral.
A: $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)
=\dfrac{1}{2}[(a+b-c)(b+c-a)(a-b)^2+(b+c-a)(a+c-b)(b-c)^2+(a+c-b)(a+b-c)(c-a)^2]
≥0$$
A: Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$ and $u\geq0$, $v\geq0$.
Hence, $\sum\limits_{cyc}(a^3b-a^2b^2)=(u^2-uv+v^2)x^2+(u^3+2u^2v-uv^2+v^3)x+2u^3v\geq0$.
Done!
A: 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 $0$
