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Sum > Sum Reciprocals (Posted on 2010-11-25) |
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A, B and C are three positive real numbers such that: A*B*C = 1, and: A + B + C > 1/A + 1/B + 1/C
Prove that precisely one (and, no more) of A, B and C is greater than 1.
Solution
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Comment 1 of 1
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(A - 1)(B - 1)(C - 1) = ABC + A + B + C - (BC + CA + AB) - 1 = A + B + C - (BC + CA + AB) since ABC = 1 = A + B + C - (1/A + 1/B + 1/C) > 0
A, B and C cannot all be greater than 1 (since ABC = 1), so it follows that exactly two of the factors in the LHS are negative and one is positive. Thus, exactly one of A, B and C is greater than 1.
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Posted by Harry
on 2010-11-26 23:21:01 |
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