Difference between revisions of "2004 USAMO Problems/Problem 5"
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== Problem 5 == | == Problem 5 == | ||
− | (''Titu Andreescu'') Let <math> | + | (''Titu Andreescu'') |
+ | Let <math>a </math>, <math>b </math>, and <math>c </math> be positive real numbers. Prove that | ||
<center> | <center> | ||
<math> | <math> | ||
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== Solutions == | == Solutions == | ||
+ | https://youtu.be/jmXSmmfO7pQ?si=dxJ6At7KHlcn2NT5 [Video Solution by little fermat] | ||
− | We first note that for positive <math> | + | 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: |
− | * Since <math> | + | * Since <math>x^2 </math> and <math>x^3 </math> must be both lesser than, both equal to, or both greater than 1, by the [[rearrangement inequality]], <math> x^2 \cdot x^3 + 1 \cdot 1 \ge x^2 \cdot 1 + 1 \cdot x^3 </math>. |
− | * Since <math> | + | * Since <math>x^2 - 1 </math> and <math>x^3 - 1 </math> have the same sign, <math> 0 \le (x^2 - 1)(x^3 - 1) = x^5 - x^3 - x^2 + 1 </math>, with equality when <math>x = 1 </math>. |
* By weighted [[AM-GM]], <math> \frac{2}{5}x^5 + \frac{3}{5} \ge x^2 </math> and <math> \frac{3}{5}x^5 + \frac{2}{5} \ge x^3 </math>. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of [[Muirhead's Inequality]]. | * By weighted [[AM-GM]], <math> \frac{2}{5}x^5 + \frac{3}{5} \ge x^2 </math> and <math> \frac{3}{5}x^5 + \frac{2}{5} \ge x^3 </math>. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of [[Muirhead's Inequality]]. | ||
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</center> | </center> | ||
− | We present two proofs of this inequality | + | We present two proofs of this inequality: |
+ | |||
+ | * By [[Hölder's Inequality]], | ||
− | |||
<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 | + | \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> | ||
− | + | We get the desired inequality by taking <math>m_{1,1} = a^3</math>, <math>m_{2,2} = b^3</math>, <math>m_{3,3} = c^3</math>, and <math>m_{x,y} = 1 </math> when <math> x \neq y </math>. We have equality if and only if <math>a = b = c = 1 </math>. | |
− | + | * Take <math>x = \sqrt{a}</math>, <math>y = \sqrt{b}</math>, and <math>z = \sqrt{c}</math>. Then some two of <math>x</math>, <math>y</math>, and <math>z</math> are both at least <math>1</math> or both at most <math>1</math>. Without loss of generality, say these are <math>x</math> and <math>y</math>. Then the sequences <math>(x, 1, 1)</math> and <math>(1, 1, y)</math> are oppositely sorted, yielding | |
<center> | <center> | ||
<math> | <math> | ||
− | + | (x^6 + 1 + 1)(1 + 1 + y^6) \ge 3(x^6 + 1 + y^6) | |
− | + | </math> | |
− | |||
− | |||
− | </math> | ||
</center> | </center> | ||
+ | by [[Chebyshev's Inequality]]. By the [[Cauchy-Schwarz Inequality]] we have | ||
+ | <center> | ||
+ | <math> | ||
+ | (x^6 + 1 + y^6)(1 + z^6 + 1) \ge (x^3 + y^3 + z^3)^2. | ||
+ | </math> | ||
+ | </center> | ||
+ | Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get | ||
+ | <center> | ||
+ | <math> | ||
+ | 3(x^3 + y^3 + z^3) \ge (x^2 + y^2 + z^2)(x+y+z), | ||
+ | </math> | ||
+ | </center> | ||
+ | and | ||
+ | <center> | ||
+ | <math> | ||
+ | (x^3 + y^3 + z^3)(x+y+z) \ge (x^2 + y^2 + z^2)^2. | ||
+ | </math> | ||
+ | </center> | ||
+ | Multiplying the above four inequalities together yields | ||
+ | <center> | ||
+ | <math> | ||
+ | (x^6 + 2)(y^6 + 2)(z^6 + 2) \ge (x^2 + y^2 + z^2)^3, | ||
+ | </math> | ||
+ | </center> | ||
+ | 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. | 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>.'' | ||
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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.