<s>[INST] 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!!!
 [/INST] 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.
</s>