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Re: If x and y are non-negative, is (x + y) greater than xy? [#permalink]
Can anybody explain a little bit more about why statement number 2 is the correct answer? I don't know what I'm doing wrong?
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Re: If x and y are non-negative, is (x + y) greater than xy? [#permalink]
Tavo wrote:
Can anybody explain a little bit more about why statement number 2 is the correct answer? I don't know what I'm doing wrong?


Hi

Can you share your approach, so that someone would be able to shed more light on this. I will try to explain from my side.

Lets consider x^2 + y^2 and also xy.
Now, (x-y)^2 = x^2 + y^2 - 2xy. (x-y)^2 will always be >= 0. So we can write:

x^2 + y^2 - 2xy >= 0 or x^2 + y^2 >= 2xy.
If x^2 + y^2 is >= 2xy, and I am sure you would agree that 2xy is greater than xy here (both x, y are non negative); then we can conclude that x^2 + y^2 > xy

So we have deduced that x^2 + y^2 is > xy and we are already given in statement 2 that (x + y) is > x^2 + y^2
Thus, (x + y) > xy
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If x and y are non-negative, is (x + y) greater than xy? [#permalink]
Given, x & y are non-negative, hence they can have any "real" value from 0 onwards.

(1) X = Y, X & Y can be = 0/1/any positive integer/any fraction. Not Sufficient.
(2) X + Y> X^2 + Y^2. Check for values 0, 1, positive integers and fractions. Condition is met only by fractions. Hence, Sufficient.

Therefore, B.
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Re: If x and y are non-negative, is (x + y) greater than xy? [#permalink]
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Re: If x and y are non-negative, is (x + y) greater than xy? [#permalink]
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