If x and y are integers, is 2xy < x^2 + y^2?

A
Becoz (x-y)^2 >0 is possible only when x not equal to y. In 2 we can not say x not equal to y but in 1 we can say.

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in S2, if xy<0 ---> means that x and y have OPPOSITE signs and are NON ZERO. Whatever be the values of x,y, you will always have x \(\neq\) y

Hope this helps.

Let's rephrase the question to be is X^2 - 2xy + Y^2 > 0, which we can break down even further to be is ABS(X-Y) > 0?.

Before we jump into each statements let's think through the rephrased question a bit more. There's only two possibilities that we need to look out for, either ABS(X-Y) = 0, or ABS(X-Y) > 0, and note that we know that ABS(X-Y) could NEVER be LESS than zero because an abs() is always >=0. So what's the only case in which ABS(X-Y) = 0? Well it's when X=Y, so in other words, we want to find whether or not X=Y.

Statement (1):
This provides us that x and y have opposite signs and neither is equal to 0. Great so remember that only one number is equal to the opposite, and that's 0, but we know from this statement that neither x or y is 0, so X must NOT be equal to Y. Sufficient.

Statement (2):
Given the restriction that x and y are integers there is never a situation in which x and y could be equal. Do note that if we were given x+y = 6 then we would have to consider the case that x=3 and y=3, but that's not the case here, so suffecient.

D is the answer.

2xy<x2+y2
0<x2-2xy+y2
0<(x-y)2
0< (x-y)
y<x

by deriving this I get option E as the answer
Can please anyone clarify, what's wrong with the equation derived?

­
All is good up to 0 < (x - y)^2, which is not equivalent to 0 < x - y. The square of a number is always non-negative; hence, 0 < (x - y)^2 is always true unless x = y. When x = y, then (x - y)^2 = 0, not greater than 0. Thus, the question essentially asks whether x = y. If x = y, then the answer to the question is NO. However, if x ≠ y, then the answer to the question is YES.

The first statement, xy < 0, implies that x and y have opposite signs and thus cannot be equal, giving a YES answer to the question and making (1) sufficient.

The second statement, x + y = 5­, is also sufficient because we are given that x and y are integers, and thus for x + y = 5 to be true, they cannot be equal.

Answer: D.

Hope it's clear.

Thanks, for statement two I also ended up writing the inequality as (x+y)^2 minus (x-y)^2 = 4xy so we have x+y = 5.

Or, we have Expression I: 25 minus 4xy = (x-y)^2;

The question is (x-y)^2 > 0? Now, Expression I 25 will always be great than or equal to 4xy as the maximum value of xy given x+y=5 is 25/4 so this statement is sufficent, maximum value occurs when x=y=2.5 which is not possible as x and y are integers.

I was not able to see that the question is asking for x=y so had to go down this route.


Thanks,
Karam