IMO 1964 Problem 2

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m53s

Problem

Suppose $a, b, c$ are the sides of a triangle. Prove that

$$ a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}. $$

Exploration

The expression is symmetric and homogeneous of degree three, suggesting a normalization or a classical symmetric inequality. Expanding each term gives a combination of cubic and mixed products, which suggests Schur-type inequalities or rearrangements involving $a^3$, $b^3$, $c^3$, and $abc$.

A direct expansion leads to

$$\sum a^2(b+c-a)=\sum a^2(b+c)-\sum a^3.$$

This reduces the problem to controlling $\sum a^2(b+c)$ in terms of $\sum a^3$ and $abc$.

A classical inequality that connects these quantities is Schur’s inequality of degree one,

$$\sum a^3 + 3abc \ge \sum a^2(b+c),$$

which matches the structure required after rearrangement.

The key difficulty lies in proving Schur’s inequality in a form compatible with triangle side conditions, preferably through an identity that expresses the difference as a sum of nonnegative terms involving $(a-b)^2$ and linear expressions in $a,b,c$.

Problem Understanding

The problem asks to prove an upper bound for a symmetric cubic expression formed from the side lengths of a triangle. The structure mixes squared terms with linear complements of the triangle sides, suggesting that the triangle inequality should be used in a multiplicative way rather than directly.

This is a Type B problem, since a fixed inequality is to be established.

The inequality should hold because the expression can be transformed into a classical symmetric form controlled by Schur’s inequality, which balances cubic terms against mixed products and the geometric constraint that each pair of sides exceeds the third.

Proof Architecture

The argument relies on a single structural identity and a positivity lemma.

The first lemma states that the expression $\sum a^2(b+c-a)$ can be rewritten as $\sum a^2(b+c)-\sum a^3$, which follows directly from algebraic expansion of each cyclic term.

The second lemma is Schur’s inequality in degree one form,

$$\sum a^3 + 3abc - \sum a^2(b+c) = \frac{1}{2}\sum (a-b)^2(a+b-c),$$

together with the fact that each term $(a-b)^2(a+b-c)$ is nonnegative for triangle side lengths since $a+b>c$ and cyclic permutations hold.

The hardest part is the factorization identity in Schur’s inequality, since an incorrect expansion or missing symmetry factor would invalidate the entire argument.

Solution

The cyclic expression expands as

$$\sum a^2(b+c-a)=\sum a^2(b+c)-\sum a^3.$$

This rewrites the problem into controlling the difference between $\sum a^2(b+c)$ and $\sum a^3$.

A structural identity holds for all real numbers $a,b,c$:

$$\sum a^3 + 3abc - \sum a^2(b+c)=\frac{1}{2}\sum (a-b)^2(a+b-c).$$

To verify this identity, expand one term:

$$(a-b)^2(a+b-c)=(a^2-2ab+b^2)(a+b-c).$$

Multiplying out gives

$$a^3 + a^2b - a^2c -2a^2b -2ab^2 +2abc + ab^2 + b^3 - b^2c.$$

Grouping like terms yields

$$a^3 + b^3 + 2abc - a^2b - ab^2 - a^2c - b^2c.$$

Summing cyclically produces

$$2\left(\sum a^3 + 3abc - \sum a^2(b+c)\right).$$

Division by $2$ gives the stated identity.

Each triangle satisfies $a+b>c$, hence each factor $a+b-c$ is positive, and each square $(a-b)^2$ is nonnegative. Every term $(a-b)^2(a+b-c)$ is therefore nonnegative, which implies

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

Rearranging yields

$$\sum a^2(b+c)-\sum a^3 \le 3abc.$$

Using the initial rewriting of the original expression gives

$$\sum a^2(b+c-a)\le 3abc.$$

This completes the proof. ∎

Verification of Key Steps

The identity relating $\sum a^3 + 3abc - \sum a^2(b+c)$ to $\frac{1}{2}\sum (a-b)^2(a+b-c)$ depends on correct cancellation of mixed terms. Recomputing from the expansion of $(a-b)^2(a+b-c)$ confirms that each term of type $a^2b$ and $ab^2$ appears twice with opposite signs across cyclic permutations, leaving exactly the required symmetric combination.

The positivity of $(a+b-c)$ depends on the triangle inequality applied to each pair of sides. Any attempt to omit the squaring factor $(a-b)^2$ would fail, since sign control of $a-b$ alone is insufficient without symmetry.

The reduction step from the original expression to $\sum a^2(b+c)-\sum a^3$ is sensitive to sign errors; expanding each cyclic term individually prevents such inconsistencies.

Alternative Approaches

A different route uses normalization $a+b+c=1$, converting the inequality into a homogeneous cubic form and applying rearrangement inequalities between ordered sequences $(a^2,a^2,a^2)$ and $(b+c-a,b+c-b,b+c-c)$. Another approach proceeds through smoothing arguments, showing that the expression increases under equalization of two variables while preserving the triangle condition, eventually reducing to the equilateral case.