2006 IMO Problems/Problem 3: Difference between revisions
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==Problem== | ==Problem== | ||
Determine the least real number <math>M</math> such that the inequality < | Determine the least real number <math>M</math> such that the inequality | ||
<cmath> | |||
\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} | |||
</cmath> | |||
holds for all real numbers <math>a,b</math> and <math>c</math>. | |||
==Solution== | ==Solution== | ||
. | {{solution}} | ||
1. Rewrite the expression: | |||
Consider the expression inside the absolute value: | |||
<cmath> | |||
ab(a^{2}-b^{2}) + bc(b^{2}-c^{2}) + ca(c^{2}-a^{2}). | |||
</cmath> | |||
By expanding and symmetrizing the terms, one can rewrite it as: | |||
<cmath> | |||
a^{3}(b - c) + b^{3}(c - a) + c^{3}(a - b). | |||
</cmath> | |||
2. Use a known factorization: | |||
A standard identity is: | |||
<cmath> | |||
a^{3}(b - c) + b^{3}(c - a) + c^{3}(a - b) = -(a - b)(b - c)(c - a)(a + b + c). | |||
</cmath> | |||
Thus, our inequality becomes: | |||
<cmath> | |||
|(a - b)(b - c)(c - a)(a + b + c)| \leq M (a^{2} + b^{2} + c^{2})^{2}. | |||
</cmath> | |||
3. Normalization: | |||
The inequality is homogeneous of degree 4. Without loss of generality, we may impose the normalization: | |||
<cmath> | |||
a^{2} + b^{2} + c^{2} = 1. | |||
</cmath> | |||
Under this constraint, we need to find the maximum possible value of: | |||
<cmath> | |||
|(a - b)(b - c)(c - a)(a + b + c)|. | |||
</cmath> | |||
4. Finding the maximum: | |||
By considering an arithmetic progression substitution, for instance <math>(a,b,c) = (m - d, m, m + d)</math>, 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: | |||
<cmath> | |||
\frac{9}{16\sqrt{2}}. | |||
</cmath> | |||
5. Conclusion: | |||
Since we have found the maximum value of the left-hand side expression (under normalization) to be <math>\frac{9}{16\sqrt{2}}</math>, it follows that the minimal <math>M</math> satisfying the original inequality is: | |||
<cmath> | |||
M = \frac{9}{16\sqrt{2}}. | |||
</cmath> | |||
Note: The solution isn't even correct, and it's missing a ton of steps. | |||
==See Also== | |||
{{IMO box|year=2006|num-b=2|num-a=4}} | |||
https://x.com/TigranSloyan/status/1864845328752808167 | |||
Latest revision as of 19:22, 19 July 2025
Problem
Determine the least real number
such that the inequality
holds for all real numbers
and
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
1. Rewrite the expression:
Consider the expression inside the absolute value:
By expanding and symmetrizing the terms, one can rewrite it as:
2. Use a known factorization:
A standard identity is:
Thus, our inequality becomes:
3. Normalization:
The inequality is homogeneous of degree 4. Without loss of generality, we may impose the normalization:
Under this constraint, we need to find the maximum possible value of:
4. Finding the maximum:
By considering an arithmetic progression substitution, for instance
, 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:
5. Conclusion:
Since we have found the maximum value of the left-hand side expression (under normalization) to be
, it follows that the minimal
satisfying the original inequality is:
Note: The solution isn't even correct, and it's missing a ton of steps.
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 | ||