# 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

• What's with the formatting on your titles? May 22, 2019 at 6:29

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

1. $$b=\max\{a,b,c\}$$, $$b=x+u+v,$$ $$a=x+u$$ 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

1. $$c=\max\{a,b,c\}$$, $$c=x+u+v,$$ $$a=x+u$$ 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: $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq b(a+b-c)(a-c)(c-b).$$

• Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
– user685500
Mar 7, 2019 at 11:16

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 $$\rfloor$$

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!

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.