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Re: If xy > 0, is y/x + x/y > 2? [#permalink]
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\(\frac{y}{x}+\frac{x}{y}>2\)
\(y^2+x^2>2xy\)
\(y^2+x^2-2xy>0\)
\((x-y)^2>0\)

From Statement 1 \(x-y=1\) -> square both sides
\((x-y)^2=1\) sufficient

From Statement 2 \(x=2y\)
\(x/y=2\)
\(y/x=1/2\)
\(\frac{x}{y}+\frac{y}{x}=2+\frac{1}{2}\)

Which is greater than 2, hence sufficient

Answer D
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Re: If xy > 0, is y/x + x/y > 2? [#permalink]
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Given, xy > 0 means neither x nor y can be ZERO
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Re: If xy > 0, is y/x + x/y > 2? [#permalink]
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Quote:
If xy>0, is y/x+x/y>2?


(1) y=x−1

(2) x=2y



If xy>0, i.e. signs of both x and y are identical
Now, y/x+x/y will be equal to 2 only if both x and y are equal otherwise it will always be greater than 2?
So question becomes

Is x = y?


(1) y=x−1

i.e. x and y are not equal hence SUFFICIENT

(2) x=2y

i.e. x and y are NOT equal hence SUFFICIENT

Answer: Option D
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Re: If xy > 0, is y/x + x/y > 2? [#permalink]
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xy > 0, So x and y will have the same sign.
We are required to find whether \(\frac{y}{x}\) + \(\frac{x}{y}\) > 2
=> \(x^2 + y^2 > 2xy\)
=> \((x + y)^2 > 4xy\) -> [a]

(1) y = x -1
=> x = y + 1
Substituting the above value in [a] we get,
\((y + 1 + y)^2 > 4 (y + 1) y\)
=> \(4y^2 + 4y + 1 > 4y^2 + 4y\)
=> \(1 > 0\) This will always be true. Hence Sufficient

(2) x = 2y
Substituting the above value in [a] we get,
\((2y + y)^2 > 4 (2y) y\)
=> \(9y^2 > 8y^2\)
=> \(9 > 8\) This will always be true. Hence sufficient

Answer D
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Re: If xy > 0, is y/x + x/y > 2? [#permalink]
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Re: If xy > 0, is y/x + x/y > 2? [#permalink]
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