Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c- b)$$
If $c\neq {\rm mid}\{a, b, c\}$, the inequality is obviously true! 
If $c={\rm mid}\{a, b, c\}$, we have $(a- c)(c- b)= 0\Leftrightarrow c= \dfrac{c^2+ ab}{a+ b}$. I tried to prove that:
$$f(c)- f(\frac{c^2+ ab}{a+ b})= (a- c)(c- b)F\geqq 0$$
where $f(c)= 3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$ but without success! I found this inequality by using discriminant and some coefficient skills. Thank you so much

The task is homogenius. Let WLOG
$$a+b+c=2,\quad a,b,c \in(0,1),\tag1$$
$$f(a,b,c) = 3\big(a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\big) - b(a+b-c)(a-c)(c-b).\tag2$$
Using of the substiutions
$$a=x,\quad b= 1-xy,\quad c=1-x+xy,\quad x,y\in(0,1),\tag3$$
provides the conditions $(1)$ and allows to get
$\quad f\big(x,1-xy,1-x+xy\big) = 2xg(x,y),$
where
\begin{cases}
g(x,y) = 3(1-y) + (-10+9y+y^2)x + (11-12y+y^2+3y^3)x^2\\[4pt]
 + (-3+4y-2y^2-y^3-y^4)x^3\\[4pt]
g(0,y) = 3(1-y)\\[4pt]
g(1,y) = 1-2y+2y^3-y^4 = (1-y)^3(1+y)\\[4pt]
g(x,0) = 3-10x+11x^2-3x^3 = (1-x)^3 + 2(1-2x)^2 + x\\[4pt]
g(x,1) = 2x^2-2x^3.\tag4
\end{cases}
From $(4)$ should $g(x,y)\ge 0$ at the edges of the area.
On the other hand, at the inner stationary points
$$4g(x,y) = 4g(x,y) - g'_x(x,y)  = 12(1-y) + 3(-10+9y+y^2)x + 2(11-12y+y^2+3y^3)x^2,$$
with the discriminant
\begin{align}
&D(y) = 9(-10+9y+y^2)^2 - 96(1-y)(11-12y+y^2+3y^3)\\[4pt]
& = -3 (1-y)(52-144y+89y^2+99y^3)\\[4pt]
& = -3 (1-y)\big(52(1-y)^3 + 12y(1-3y)^2 + 5y^2+43y^3\big) < 0.
\end{align}
Therefore, $g(x,y) \ge 0.$
$\color{brown}{\textbf{Is proved.}}$
$\color{green}{\textbf{Notes about the areas.}}$
The area $c=\operatorname{med}(a,b,c),\quad c\in\big[\min(a,b),\max(a,b)\big],$ corresponds with the area
$$y\in\left(\min\left(\frac12,2-\dfrac1x\right),\max\left(\frac12,2-\dfrac1x\right)\right)$$
(the plot of the area bounds see below).

However, applied universal approach allows to avoid such detalization.
