Difference between revisions of "2004 USAMO Problems/Problem 5"
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− | \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} | + | \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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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 12:58, 27 August 2023
Problem 5
(Titu Andreescu)
Let ,
, and
be positive real numbers. Prove that
.
Solutions
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.