Difference between revisions of "1993 USAMO Problems/Problem 1"
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== See also == | == See also == | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Revision as of 14:29, 15 April 2012
Problem
For each integer , determine, with proof, which of the two positive real numbers and satisfying is larger.
Solution
Square and rearrange the first equation and also rearrange the second. It is trivial that since clearly cannot equal (Otherwise ). Thus where we substituted in equations (1) and (2) to achieve (5). Notice that from we have . Thus, if , then . Since , multiplying the two inequalities yields , a contradiction, so . However, when equals or , the first equation becomes meaningless, so we conclude that for each integer , we always have .
See also
1993 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |