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

I found a nice identity to prove this!
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$$
$$=(a + b - c)(a + c)(a - c)^2 + (a + b - c)( c + b-a)(a - b)^2 + ( c + b-a)(2\,a - b + c)( b-c)^2 \geqq 0$$
By the way$,$ with $k=constant, k \in [0,1]$ and $a,b,c$ is three side of the triangle$:$
$$\sum\,\it{a}^{\,\it{2}}\it{b}\it{(}\,\,\it{a}- \it{b}\,\,\it{)}\geqq \it{k}\,.\,\it{b}\it{(}\,\,\it{a}+ \it{b}- \it{c}\,\,\it{)}\it{(}\,\,\it{a}- \it{c}\,\,\it{)}\it{(}\,\,\it{c}- \it{b}\,\,\it{)}$$
Proof: $$\text{LHS}-\text{RHS}=k \left\{ b \left( a+b-c \right)  \left( a-c \right) ^{2}+a \left( b+c- a \right)  \left( b-c \right) ^{2} \right\} + \left( 1-k \right) \left\{ {a}^{2}b \left( a-b \right) +{b}^{2}c \left( -c+b \right) +{c} ^{2}a \left( -a+c \right)  \right\}$$
Where the last inequality$:$ $$ {a}^{2}b \left( a-b \right) +{b}^{2}c \left( -c+b \right) +{c} ^{2}a \left( -a+c \right)  \geqq 0$$ is IMO 1983!
You can see also here.
