1993 USAMO Problems/Problem 1
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). 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 .
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