2004 USAMO Problems/Problem 5
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 |
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