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Zorluk: OrtaQuestion Stem Simplification and Target Rephrasing

For all non-zero real numbers xx and yy such that x≠yx \neq y, the Data Sufficiency Yes/No target question "Is xy+yx>2\frac{x}{y} + \frac{y}{x} > 2?" is algebraically equivalent to the simplified target question "Is xy>0xy > 0?"

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True
The statement is True because rewriting the inequality as (x−y)2xy>0\frac{(x - y)^2}{xy} > 0 shows that since (x−y)2>0(x - y)^2 > 0 for all x≠yx \neq y, the inequality holds if and only if xy>0xy > 0.

Adım Adım Çözüm

1
Combine the fractions on the left-hand side of the target inequality
x2+y2xy>2\frac{x^2 + y^2}{xy} > 2
Finding a common denominator xyxy combines the two reciprocal ratios into a single rational expression.
2
Move all terms to one side to compare against zero
x2+y2xy−2>0  ⟹  x2−2xy+y2xy>0\frac{x^2 + y^2}{xy} - 2 > 0 \implies \frac{x^2 - 2xy + y^2}{xy} > 0
Comparing rational expressions directly to zero simplifies testing for positive or negative regions.
3
Factor the numerator into a perfect square
(x−y)2xy>0\frac{(x - y)^2}{xy} > 0
Recognizing x2−2xy+y2x^2 - 2xy + y^2 as (x−y)2(x - y)^2 isolates a term with a guaranteed non-negative property.
4
Evaluate sign conditions under the problem constraints
Because x≠yx \neq y, (x−y)2(x - y)^2 is strictly positive. Thus, (x−y)2xy>0\frac{(x - y)^2}{xy} > 0 holds if and only if xy>0xy > 0.
A quotient with a strictly positive numerator is positive if and only if its denominator is positive.

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Question Stem Simplification via Rational Expression Factoring
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Question Stem Simplification and Target Rephrasing

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