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} | |
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Revision as of 23:54, 1 July 2008
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 .
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.