Difference between revisions of "1993 USAMO Problems/Problem 1"
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Revision as of 14:08, 22 March 2008
For each integer , determine, with proof, which of the two positive real numbers and satisfying is larger.
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). If , then since , , and are all positive. Adding the two would mean , a contradiction, so . However, when equals or , the first equation becomes meaningless, so we conclude that for each integer , we always have .
|1993 USAMO (Problems • Resources)|
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