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Difficulty: Very hardQuestion Stem Simplification and Target Rephrasing

For all non-zero real numbers mm and nn with m≠nm \neq n, the Data Sufficiency target question "Is m2+n22mn>1\frac{m^2 + n^2}{2mn} > 1?" is algebraically equivalent to the simplified target question "Are mm and nn of opposite signs?".

Answer: Answer

Answer

False. The target question "Is m2+n22mn>1\frac{m^2 + n^2}{2mn} > 1?" simplifies to "Do mm and nn have the same sign?", not opposite signs.
The statement is false. Rephrasing m2+n22mn>1\frac{m^2 + n^2}{2mn} > 1 yields (m−n)22mn>0\frac{(m-n)^2}{2mn} > 0. Because (m−n)2>0(m-n)^2 > 0 when m≠nm \neq n, the ratio is positive if and only if mn>0mn > 0, which requires mm and nn to have the same sign, not opposite signs.

Step-by-Step Solution

1
Subtract 1 from both sides of the inequality to consolidate into a single rational expression.
m2+n22mn−1>0  ⟹  m2−2mn+n22mn>0\frac{m^2 + n^2}{2mn} - 1 > 0 \implies \frac{m^2 - 2mn + n^2}{2mn} > 0
Consolidating terms onto one side avoids cross-multiplying by 2mn2mn, which has an unknown sign.
2
Factor the numerator quadratic expression.
(m−n)22mn>0\frac{(m - n)^2}{2mn} > 0
Recognize that m2−2mn+n2m^2 - 2mn + n^2 is a perfect square binomial (m−n)2(m - n)^2.
3
Determine the sign conditions of the numerator and denominator.
Since m≠nm \neq n, (m−n)2>0(m - n)^2 > 0 is strictly positive. Therefore, (m−n)22mn>0\frac{(m - n)^2}{2mn} > 0 holds if and only if 2mn>0  ⟹  mn>02mn > 0 \implies mn > 0.
A fraction with a positive numerator is strictly greater than zero if and only if its denominator is also strictly positive.
4
Translate the condition mn>0mn > 0 into verbal sign properties.
mn>0mn > 0 implies that mm and nn share the SAME non-zero sign (both positive or both negative).
Comparing this condition to the statement in the prompt reveals that the claim of 'opposite signs' is incorrect.

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Question Stem Simplification and Target Rephrasing
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Question Stem Simplification and Target Rephrasing

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