Given triangle side lengths $a, b, c$, show that $3\left(a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\right)\geq b\left(a+b-c\right)\left(a-c\right)\left(c-b\right)$

Question

If you're interested in IMO 1983-style inequalities, please consider the following problem:

Given three positive real numbers $a, b, c$ that form the side lengths of a triangle, prove inequality: $$3\left ( a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ) \right )\geq b\left ( a+ b- c \right )\left ( a- c \right )\left ( c- b \right )$$

• If $c\neq\operatorname{med}\left ( a, b, c \right )$, then the inequality is clearly true by sign analysis.
• The critical case is when $c= \operatorname{med}\left ( a, b, c \right )$, i.e., when $c$ is the median among the three values.

In this case, $\left ( a- c \right )\left ( c- b \right )= 0$, and that occurs when: $$c= \frac{c^{2}+ ab}{a+ b}$$ To address this case, I attempted to prove the identity: $$f\left ( c \right )- f\left ( \frac{c^2+ ab}{a+ b} \right )= \left ( a- c \right )\left ( c- b \right )\cdot F\geq 0$$ where $$f\left ( c \right )= 3\left ( a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ) \right )- b\left ( a+ b- c \right )\left ( a- c \right )\left ( c- b \right )$$ However, I haven’t succeeded in completing the proof. I discovered this inequality using discriminant analysis and coefficient comparison techniques.

Answer 1

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

Answer 2

To use the Ravi substitution, we let: $a= y+ z, b= z+ x, c= x+ y$. Then the problem becomes:

For $x, y, z> 0$, we need to prove the inequality: $$3x^{3}z- 2x^{2}yz- x^{2}z^{2}+ 3xy^{3}- 3xy^{2}z- 3xyz^{2}+ 2yz^{3}+ z^{4} \geq 0$$ $$\because\quad 3x^{3}z- 2x^{2}yz- x^{2}z^{2}+ 3xy^{3}- 3xy^{2}z- 3xyz^{2}+ 2yz^{3}+ z^{4}- z\left ( x+ 2y+ z \right )\left ( z- x \right )^{2}\geq 0$$ $$\because yz\left ( 2x^{2}- 4xy+ 3y^{2}- 2yz+ z^{2} \right )+ 3xy\left ( y- z \right )^{2}\geq 0\because 2x^{2}- 4xy+ 3y^{2}- 2yz+ z^{2}\geq 0.$$ This can be rewritten using the S.O.S. (Sum of Squares) method: $$\begin{array}{rl} 2x^{2}- 4xy+ 3y^{2}- 2yz+ z^{2} & = \left ( 2x- y- z \right )^{2}- 2\left ( x^{2}- 2xz- y^{2}+ 2yz \right )\\ & = \left ( x- 2y+ z \right )^{2}+ \left ( x^{2}- 2xz- y^{2}+ 2yz \right )\\ & \geq 0.\\ \end{array}$$

Answer 3

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.

Answer 4

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.