Difference between revisions of "2004 USAMO Problems/Problem 5"
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== Solutions == | == Solutions == | ||
+ | https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat] | ||
We first note that for positive <math>x </math>, <math> x^5 + 1 \ge x^3 + x^2 </math>. We may prove this in the following ways: | We first note that for positive <math>x </math>, <math> x^5 + 1 \ge x^3 + x^2 </math>. We may prove this in the following ways: | ||
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<center> | <center> | ||
<math> | <math> | ||
− | + | \begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \ \qquad \qquad \qquad\left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3.\end{matrix} | |
</math> | </math> | ||
</center> | </center> | ||
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</center> | </center> | ||
as desired, with equality if and only if <math>x = y = z = 1</math>. | as desired, with equality if and only if <math>x = y = z = 1</math>. | ||
+ | * First, expand the left side of the inequality to get | ||
+ | <center> | ||
+ | <math>a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8.</math> | ||
+ | </center> | ||
+ | By the AM-GM inequality, it is true that <math>a^3 + a^3b^3 + 1 \ge 3a^2b</math>, and so it is clear that | ||
+ | <center> | ||
+ | <math>a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8 \ge a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2.</math> | ||
+ | </center> | ||
+ | Additionally, again by AM-GM, it is true that <math>a^3b^3c^3 + a^3 + b^3 + c^3 + 1 + 1 \ge 6abc</math>, and so | ||
+ | <center> | ||
+ | <math>a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2</math> | ||
+ | </center> | ||
+ | |||
+ | <center> | ||
+ | <math>\ge a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2 + 6abc = {(a + b + c)}^3,</math> | ||
+ | </center> | ||
+ | as desired. | ||
+ | |||
+ | |||
''It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that <math>x^5 - x^2 + 3 \ge x^3 + 2 </math>.'' | ''It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that <math>x^5 - x^2 + 3 \ge x^3 + 2 </math>.'' | ||
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== Resources == | == Resources == | ||
− | + | {{USAMO newbox|year=2004|num-b=4|num-a=6}} | |
− | + | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:19, 26 September 2024
Problem 5
(Titu Andreescu) Let , , and be positive real numbers. Prove that
.
Solutions
https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat]
We first note that for positive , . We may prove this in the following ways:
- Since and must be both lesser than, both equal to, or both greater than 1, by the rearrangement inequality, .
- Since and have the same sign, , with equality when .
- By weighted AM-GM, and . Adding these gives the desired inequality. Equivalently, the desired inequality is a case of Muirhead's Inequality.
It thus becomes sufficient to prove that
.
We present two proofs of this inequality:
We get the desired inequality by taking , , , and when . We have equality if and only if .
- Take , , and . Then some two of , , and are both at least or both at most . Without loss of generality, say these are and . Then the sequences and are oppositely sorted, yielding
by Chebyshev's Inequality. By the Cauchy-Schwarz Inequality we have
Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get
and
Multiplying the above four inequalities together yields
as desired, with equality if and only if .
- First, expand the left side of the inequality to get
By the AM-GM inequality, it is true that , and so it is clear that
Additionally, again by AM-GM, it is true that , and so
as desired.
It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2004 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.