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# If x and y are non-negative, is (x + y) greater than xy?

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

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07 Feb 2018, 21:21
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If x and y are non-negative, is (x + y) greater than xy?

(1) x = y

(2) x + y is greater than x^2 + y^2

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

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07 Feb 2018, 22:02
If x and y are non-negative, is (x + y) greater than xy?

(1) x = y
If x=y, the question stem becomes $$2x > x^2$$

If x=y=$$\frac{1}{2}$$, x+y is greater than xy
If x=y=5, x+y is less than xy (Insufficent)

(2) x + y is greater than x^2 + y^2

If x=y=$$\frac{1}{2}$$, x+y is greater than xy
If x=$$\frac{1}{2}$$, y=$$\frac{3}{4}$$, x+y is greater than xy (Sufficient - Option B)
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Re: If x and y are non-negative, is (x + y) greater than xy?  [#permalink]

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08 Feb 2018, 04:49
2
1
Bunuel wrote:
If x and y are non-negative, is (x + y) greater than xy?

(1) x = y

(2) x + y is greater than x^2 + y^2

As we're not asked about general logical properties (such as positive/negative), well go for an equation-driven approach.
This is a Precise methodology.

We'll write an 'inequality' with a question mark:
$$x + y (?) xy$$

(1) Substituting into our question stem gives $$2x (?) x^2$$ --> $$0 (?) x^2 - 2x$$ --> $$0 (?) x(x - 2).$$
Since we don't know if $$x(x - 2)$$ is positive or not, this is insufficient.

(2) gives $$x + y > x^2 + y^2$$. Looking at this expression we should be reminded of the core identity: $$(x + y)^2 = x^2 + 2xy + y^2$$
Using this, we can write:
$$x + y - 2xy > x^2 + y^2 - 2xy$$ which gives $$x + y - 2xy > (x - y)^2$$ and then $$x + y > (x - y)^2 + 2xy$$
This essentially says that (x + y) is larger than a positive number plus 2xy.
Then (x + y) is larger than 2xy meaning it is larger than xy.
Sufficient!

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

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20 Feb 2018, 20:50
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]

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20 Feb 2018, 22:58
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]

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21 Feb 2018, 09:31
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.
If x and y are non-negative, is (x + y) greater than xy?   [#permalink] 21 Feb 2018, 09:31
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# If x and y are non-negative, is (x + y) greater than xy?

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