A good combination of an Olympiad inequality and its solution

Question

Given three side-lengths $a, b, c$ of a triangle. Prove that $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq\left | \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c \right |$$ Source: StackMath/@haidangel

buffalo-way kills it, however, I'm looking forward to another solution that dealed the inverse type of Bernhard Leeb's formula with $b:={\rm med}\left ( a, b, c \right )$ $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ):= \left ( c+ a- b \right )\left ( c- a \right )^{2}b- \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c$$

Answer 1

In the book Over and Over Again by C. Chang and T. W. Sederberg, page 15 ends with this

PROBLEM 3.2 [IMO 1983/6]. Let $a,b$, and $c$ be the lengths of the sides of a triangle. Prove that $$ (3.7) \qquad \qquad a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0. $$

Continuing later on page 16 is this

We mention here an interesting story of the IMO in 1983. Bernhard Leeb, a team member of the Federal Republic of Germany, was awarded the special prize of that year for his "one line proof" of (3.7). Leeb pointed out that $(3.7)$ has an equivalent form $$ (3.10) \quad a(b-c)^2(b+c-a)+b(a-b)(a-c)(a+b-c)\ge 0. $$ Note that the first term in (3.10) is always nonnegative and that a cyclic permutation of $\,(a,b,c)\,$ leaves the given quantity (3.7) unchanged, so we can assume without loss of generality that $\,a\,$ is the maximal side length. Thus the second term in (3.10) is nonnegative too. Equality holds only if $\,a=b=c.$

Use the triangle inequality to define new values $$ x \!:=\! b\!+\!c\!-\!a\!>\!0, \quad y \!:=\! c\!+\!a\!-\!b\!>\!0, \quad z \!:=\! a\!+\!b\!-\!c\!>\!0. \tag{1} $$

Define the quantity we are interested in $$ Q :=a^2b(a-b)+ b^2c(b-c)+ c^2a(c-a). \tag{2} $$ The Bernhard Leeb identity states that $$ Q = a(b-c)^2(b+c-a) + b(a-b)(a-c)(a+b-c). \tag{3} $$ The quantity $\,Q\,$ is invariant if $\,a,b,c\,$ is permuted cyclically. Thus, equation $(3)$ implies $$ Q = (c+a-b)(c-a)^2b - (b+c-a)(a-b)(b-c)c. \tag{4} $$ Apply the definitions in $(1)$ to equation $(4)$ to get $$ Q = y(c-a)^2b - x(a-b)(b-c)c. \tag{5} $$ Suppose the side lengths are ordered so that $\, c>a>b>0. \,$ This implies $\,a-b>0,\,b-c<0.\,$ Combined with $\,x>0,y>0\,$ applied to equation $(5)$ we get $$ Q > y(c-a)^2b > 0\quad \text{ and }\quad Q > |x(a-b)(b-c)c| > 0. \tag{6} $$

Note that if we name the largest side as $\,c\,$ and if we name the other two sides so that $\,c>b>a>0\,$ instead, then the second inequality in equation $(6)$ need not hold.

The simplest example of such a reverse inequality is $$ a\!=\!2,b\!=\!3,c\!=\!4 \;\; \text{ where }\;\; Q\!=\!16 \!<\! x(a-b)(b-c)c \!=\!20. \tag{7} $$