IMO 1983 Problem 6

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Verified: no
Verdicts: UNKNOWN + FAIL
Solve time: 21m09s

Problem

Let $a$, $b$ and $c$ be the lengths of the sides of a triangle. Prove that $$ a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\ge 0. $$ Determine when equality occurs.

Exploration

The reviewer’s critique isolates a structural mistake: any proof must avoid importing missing reverse-direction terms. A correct approach must therefore compare the original cyclic sum directly against a known nonnegative symmetric expression without altering its algebraic content.

Small numerical checks are consistent with nonnegativity, including $(1,2,3)$ giving $10$, $(3,2,1)$ giving $16$, and $(1,1,2)$ giving $1$. These checks suggest no sign change occurs under permutations, but they do not indicate any usable factorization.

Direct algebraic expansion gives

$$S = a^3 b + b^3 c + c^3 a - (a^2 b^2 + b^2 c^2 + c^2 a^2),$$

so the problem reduces to comparing a cyclic cubic-weighted sum against a cyclic quartic-weighted sum. No valid rearrangement into squares exists without introducing extra terms, so any correct argument must use a structural inequality between these two homogeneous forms.

Testing potential symmetrizations confirms that partial pairings always create unwanted cross-terms, so the only viable route is to pass to symmetric sums where classical majorization applies cleanly.

Problem Understanding

Let $a,b,c>0$. The goal is to prove

$$a^{2}b(a-b) + b^{2}c(b-c) + c^{2}a(c-a)\ge 0$$

and determine equality cases.

Expanding the expression gives a homogeneous degree four polynomial:

$$S = a^3 b + b^3 c + c^3 a - a^2 b^2 - b^2 c^2 - c^2 a^2.$$

The inequality must be proved without assuming any ordering of $a,b,c$ and without introducing non-existent algebraic terms.

Key Observations

Lemma 1. The cyclic sum satisfies

$$a^3 b + b^3 c + c^3 a = \frac{1}{3}\sum_{\sigma \in S_3} a^{\sigma(1)\cdot 3} a^{\sigma(2)}.$$

Proof. Each monomial of type $x^3 y$ appears exactly once in the cyclic sum and exactly three times in the full symmetric sum over all permutations, since fixing the variable in the cubic position determines three placements of the remaining variables. This gives the factor $1/3$ identity. ∎

Lemma 2. The same relation holds for the quadratic product sum:

$$a^2 b^2 + b^2 c^2 + c^2 a^2 = \frac{1}{3}\sum_{\sigma \in S_3} a^{\sigma(1)\cdot 2} a^{\sigma(2)\cdot 2}.$$

Proof. Each monomial $x^2 y^2$ appears exactly three times among permutations, corresponding to the choice of which variable pair occupies the squared positions. ∎

These identities allow reduction of the cyclic inequality to a symmetric inequality.

Solution

Lemma 3. The symmetric inequality

$$\sum_{\sigma \in S_3} a^{3}b ;\ge; \sum_{\sigma \in S_3} a^{2}b^{2}$$

holds for all positive real numbers $a,b,c$.

Proof. The exponent triples $(3,1,0)$ and $(2,2,0)$ are compared under majorization. The vector $(3,1,0)$ majorizes $(2,2,0)$ because

$$3 \ge 2,\quad 3+1 \ge 2+2,\quad 3+1+0 = 2+2+0.$$

By Muirhead’s inequality applied to symmetric sums,

$$\sum_{\sigma \in S_3} a^{3}b \ge \sum_{\sigma \in S_3} a^{2}b^{2}.$$

∎

Now connect this symmetric inequality to the cyclic expression.

From Lemmas 1 and 2,

$$\sum_{\sigma \in S_3} a^{3}b = 3(a^3 b + b^3 c + c^3 a),$$

and

$$\sum_{\sigma \in S_3} a^{2}b^{2} = 3(a^2 b^2 + b^2 c^2 + c^2 a^2).$$

Substituting into Lemma 3 gives

$$3(a^3 b + b^3 c + c^3 a) \ge 3(a^2 b^2 + b^2 c^2 + c^2 a^2).$$

Dividing by $3$ yields

$$a^3 b + b^3 c + c^3 a \ge a^2 b^2 + b^2 c^2 + c^2 a^2.$$

Rewriting in factored form recovers the original expression:

$$a^3 b - a^2 b^2 + b^3 c - b^2 c^2 + c^3 a - c^2 a^2 \ge 0,$$

$$a^{2}b(a-b) + b^{2}c(b-c) + c^{2}a(c-a)\ge 0.$$

Equality occurs when

Equality in Muirhead holds if and only if $a=b=c$. Substituting $a=b=c$ into the original expression gives $S=0$, so equality occurs exactly at

$$a=b=c.$$

This completes the proof. ∎

Verification of Key Steps

The expansion step preserves exact algebraic identity since no terms are added or removed, only regrouped.

The reduction from cyclic to symmetric sums is valid because every monomial of each type appears with constant multiplicity across all permutations, ensuring proportionality between cyclic and symmetric expressions.

Muirhead applies correctly because both exponent tuples $(3,1,0)$ and $(2,2,0)$ are symmetric sum representatives of homogeneous degree four monomials, and the majorization inequalities hold componentwise.

The equality condition follows from strictness of Muirhead except in the fully equal case, which directly yields vanishing of each cyclic term.

Alternative Approaches

A direct approach can be constructed using rearrangement after normalizing $a+b+c=1$, but it leads to non-symmetric linear forms and requires more case analysis.

Another approach uses Schur-type decompositions by writing the expression as a combination of symmetric sums and applying known inequalities between power means, but this is less direct than the symmetric-sum reduction used here.