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Given three side lengths $a, b, c$ of a triangle, prove the inequality: $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq\left | \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c \right |$$ Consider the identity: $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )= \left ( c+ a- b \right )\left ( c- a \right )^{2}b- \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c.$$ When $b\neq\operatorname{med}\left ( a, b, c \right )$, the left-hand side is positive.

Can you prove the inequality using this identity when $b= \operatorname{med}\left ( a, b, c \right )$? Thank you!

Dang Dang
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    Is $:=$ used for definition usually? You second equation is an identity. Maybe $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \equiv (c + a - b)(c-a)^2b - (b + c - a)(a-b)(b-c)c$ or just $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) = (c + a - b)(c-a)^2b - (b + c - a)(a-b)(b-c)c$? Similarly, in AoPS/MSE, WLOG, assume $a = \max(a, b, c)$ is commonly used rather than $a := \max(a, b, c)$? – River Li Mar 07 '21 at 09:41
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    @haidangel You may need to give information about Bernhard Leeb's formula, e.g. any other example of application of the inverse type of Bernhard Leeb's formula? – River Li Mar 07 '21 at 10:10
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    Congratulations! Would you edit your question? – River Li Mar 12 '21 at 01:52
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    Where did you see Bernhard Leeb's formula? AoPS? – River Li Mar 22 '21 at 14:42
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    https://artofproblemsolving.com/community/q1h55327p13511931 Bernhard Leeb received a special prize for rewriting the inequality as follows: Let $a = \max(a, b,c)$. $a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)=a(b+c-a)(b-c)^2+b(a+b-c)(a-b)(a-c)\geq 0$. What do you mean "inverse type"? – River Li Mar 22 '21 at 15:25
  • This is very much related to MSE question 4049630 "The inverse type of Bernhard Leeb's solution for IMO‐1983–inequality". – Somos Mar 28 '21 at 12:12

1 Answers1

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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 $$ a\!=\!2,b\!=\!3,c\!=\!4 \;\; \text{ where }\;\; Q\!=\!16 \!<\! x(a-b)(b-c)c \!=\!20. \tag{7} $$

Somos
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