Examkin

Question

Difficulty: MediumQuestion 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?"

Answer: Answer

Answer

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.

Step-by-Step Solution

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.

Key Concept

Question Stem Simplification via Rational Expression Factoring
Rate this question

Topics

Question Stem Simplification and Target Rephrasing

More questions from this topic