If $a, b, c$ are the sidelengths of a triangle, show that $a^2b(a−b) +b^2c(b−c)+c^2a(c−a)\ge0$. After many days of work and some help with some helpful Math Stack Exchange community members, I have only one inequality homework question which remains unsolved:

If $a, b, c$ are the sidelengths of a triangle, show that $a^2b(a−b) +b^2c(b−c)+c^2a(c−a)\ge0$.

My attempt:
Let $a=y+z, b=z+x, c=x+y$. Then $x,y,z\ge0$.
But after substitute into the inequality, and expand, I still cannot use Muirhead.
This question is one of the starred question and I can't do it. 
Can someone help me? Any help is appreciated!
 A: Using the Ravi-substitution we get
$$xy^3+x^3z+yz^3\geq xyz(x+y+z)$$ or
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z$$
Now using Cauchy Schwarz in Engelform:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{(x+y+z)^2}{x+y+z}=x+y+z.$$
A: Let $c=\min\{a,b,c\}$.
Thus, we need to prove that
$$\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\geq\frac{a^2b^2+a^2c^2+b^2c^2}{abc}$$ or
$$\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{a^2}{c}-\frac{a^2}{b}+\frac{c^2}{b}-c\geq\frac{a^2b^2+a^2c^2+b^2c^2-abc(a+b+c)}{abc}$$ or
$$\frac{(a-b)^2(a+b)}{ab}+\frac{(c^2-a^2)(c-b)}{bc}\geq\frac{(ac-bc)^2+(bc-ab)(ac-ab)}{abc}$$ or
$$\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c-b)(a+c)}{bc}\geq\frac{(a-b)^2c}{ab}+\frac{(c-a)(c-b)}{c}.$$ 
Can you end it now?
A: Hint
Schur inequality might be helpful
$$\sum_{cyc}a^2(a-b)(a-c) \geq 0$$
Now, the triangle inequalities like $a-c \leq b \Leftrightarrow a \leq b+c$ might be helpful.
A: The same inequality holds for bigger exponents i.e:
$$\sum_{\text{cyc}}a^pb(a-b)\geq 0$$
for $p\geq 2$ and $a,b,c$ sides of a triangle. You can prove it easily by thinking of the expression as $f(p)$, a function of $p$, and take a derivative to show $f(p)$ is non-decreasing and so 
$$f(p)\geq f(2)\geq 0.$$ 
