2006 IMO Problems/Problem 3

Problem

Determine the least real number $M$ such that the inequality $\left| ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})\right|\leq M(a^{2}+b^{2}+c^{2})^{2}$ holds for all real numbers $a,b$ and $c$ .

Solution

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1. Rewrite the expression:

Consider the expression inside the absolute value: $ab(a^{2}-b^{2}) + bc(b^{2}-c^{2}) + ca(c^{2}-a^{2}).$

By expanding and symmetrizing the terms, one can rewrite it as: $a^{3}(b - c) + b^{3}(c - a) + c^{3}(a - b).$

2. Use a known factorization:

A standard identity is: $a^{3}(b - c) + b^{3}(c - a) + c^{3}(a - b) = -(a - b)(b - c)(c - a)(a + b + c).$

Thus, our inequality becomes: $|(a - b)(b - c)(c - a)(a + b + c)| \leq M (a^{2} + b^{2} + c^{2})^{2}.$

3. Normalization:

The inequality is homogeneous of degree 4. Without loss of generality, we may impose the normalization: $a^{2} + b^{2} + c^{2} = 1.$

Under this constraint, we need to find the maximum possible value of: $|(a - b)(b - c)(c - a)(a + b + c)|.$

4. Finding the maximum:

By considering an arithmetic progression substitution, for instance $(a,b,c) = (m - d, m, m + d)$ , and analyzing the resulting expression, it can be shown through careful algebraic manipulation and optimization that the maximum value under the unit norm constraint is: $\frac{9}{16\sqrt{2}}.$

5. Conclusion:

Since we have found the maximum value of the left-hand side expression (under normalization) to be $\frac{9}{16\sqrt{2}}$ , it follows that the minimal $M$ satisfying the original inequality is: $M = \frac{9}{16\sqrt{2}}.$

2006 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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2006 IMO Problems/Problem 3

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