2006 IMO Problems/Problem 3

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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: \[M = \frac{9}{16\sqrt{2}}.\]


See Also

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

https://x.com/TigranSloyan/status/1864845328752808167