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

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    $\begingroup$ What's with the formatting on your titles? $\endgroup$ May 22, 2019 at 6:29

4 Answers 4


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: $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq b(a+b-c)(a-c)(c-b).$$

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    $\begingroup$ Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$ $\endgroup$
    – 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!

You can see also here.


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).

Area c between a and b.

However, applied universal approach allows to avoid such detalization.


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