QuantMadeEasy wrote:
Is \(x^2 + y^2 – xy >0?\)
1) xy≠0
2) xy>0
Imagine you have a right triangle, with short sides of lengths x and y. We'd find the hypotenuse using Pythagoras; it is √(x^2 + y^2) long. That hypotenuse is longer than x and longer than y. So if we make a rectangle using the sides x and y, its area, xy, will automatically be smaller than the area of the square we'd get by using the hypotenuse as one side (because each side of the rectangle is shorter than a side of the square). But that square's area is x^2 + y^2.
So it's always true, when x and y can be lengths, that x^2 + y^2 > xy. That inequality is also clearly true when xy is negative, since then the right side is negative and the left is positive (and when both x and y are negative, that's just like when they're both positive). The only situation where the inequality in the question can be false is when xy = 0. Since each Statement rules out that possibility, the answer is D.
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