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

Consider three cases.

1. $a=\max\{a,b,c\}$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$ $$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)\geq0;$$

2. $b=\max\{a,b,c\}$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$ $$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)\geq0$$ and

3. $c=\max\{a,b,c\}$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $u\geq0$ and $v\geq0.$

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$ $$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3v\geq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that: $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq b(a+b-c)(a-c)(c-b).$$

To use $\lceil$ RAVI-substitution $\rfloor$. To let $a= y+ z, b= z+ x, c= x+ y$, the problem will be become: For $x,\!y,\!z>\!0$, we need to prove $3x^{3}z- 2x^{2}yz- x^{2}z^{2}+ 3\,xy^{3}- 3xy^{\,2}z- 3xyz^{2}+ 2yz^{3}+ z^{4} \geqq 0$ $$\because\,3x^{\,3}z- 2x^{\,2}yz- x^{\,2}z^{\,2}+ 3xy^{\,3}- 3xy^{\,2}z- 3xyz^{\,2}+ 2yz^{\,3}+ z^{\,4}- z(\!x+ 2y+ z\!)(\!z- x\!)^{\,2} \geqq 0$$ $$\because yz(\!2\,x^{\,2}- 4\,xy+ 3\,y^{\,2}- 2\,yz+ z^{\,2}\!)+ 3\,xy(\!y- z\!)^{\,2}\geqq 0\because 2\,x^{\,2}- 4\,xy+ 3\,y^{\,2}- 2\,yz+ z^{\,2} \geqq 0$$ We can use $\lceil$ DRIVE!S.O.S $\rfloor$ and the following equalities. It also can be written as two squares from $$2x^{\!2}\!-\!4xy\!+\!3y^{\!2}\!-\!2yz\!+\!z^{\!2}\!=\!(\!2x\!-\!y\!-\!z\!)^{\!2}\!-\!2(\!x^{\!2}\!-\!2xz\!-\!y^{\!2}\!+\!2yz\!)\!=\!(\!x\!-\!2y\!+\!z\!)^{\!2}\!+\!(\!x^{\!2}\!-\!2xz\!-\!y^{\!2}\!+\!2yz\!)$$ q.e.d. You can also see here $\lceil$ https://h-a-i-d-a-n-g-e-l.hatenablog.com/entry/2019/03/10/200927 $\rfloor$