If x ≠ y, is 1/(x - y) < y - x?

The red part is not correct. You cannot multiply both sides of the inequality by \(x-y\) as you don't know whether this expression is positive or negative. Also answer to the question "is \((x - y)^2 < -1\)?" is always NO as the square of a number can not be negative. Basically if \(x-y>0\) the answer is always NO and if \(x-y<0\) the answer is always YES.

You can simplify the question as follows: is \(\frac{1}{x-y}<y-x\)? --> is \(y-x+\frac{1}{y-x}>0\)? is \(\frac{(y-x)^2+1}{y-x}>0\)? As the nominator (\((y-x)^2+1\)) is always positive then the question basically becomes whether denominator (\(y-x\)) is positive too --> is \(y-x>0\)? or is \(y>x\)?

You are right, thanks for pointing that.

\(1/(x-y) < (y-x)\)?
rearranging,
\((1/(x-y)) + (x-y) < 0\) ?

if \(x > y\), then \((1/(x-y)) + (x-y)\) is positive
if \(x < y\), then \((1/(x-y)) + (x-y)\) is negative,
so question becomes \(x < y\) ?

Statement 1:
\(x^2 + y^2 < 4xy\)
=>\(x^2 + y^2 - 2xy < 2xy\)
=> \((x - y)^2 < 2xy\)
since \((x - y)^2\) is greater than zero,
so \(2xy > 0\), but this does not tell whether \(x < y\) => Insuff

Statement 2: Directly answers the question => suff

Answer (B)

I had a quick question on statement 1. We have derived that (x - y) ^ 2 > 0. Why cannot we write this as (x - y) >0 and thus x > y. Hence making statement 1 as sufficient. Can someone explain why we cannot use this approach, instead choose 2xy> "some positive number"

Not sure I can follow. First of all, where and how was it derived that (x - y)^2 > 0? Also, (x - y)^2 > 0 for all values except when x = y.

Thanks for your reply Bunuel.
Let me elaborate:
In statement 1, we have:
x^2 + y^2 < 4xy
=> x^2 + y^2 - 2xy < 2xy
=> (x - y) ^ 2 < 2xy.

From here, hellosantosh2k2 chooses 2xy is positive and continues to solve the problem and arrive that statement is 1 insufficient.

At this point I wanted to take an alternate route of choosing (x - y)^2 as positive and explore that term to check for sufficiency.

By explanation and my question is below:
We can say that (x - y)^ 2 is positive as square of (x - y) will always be positive. It can also take a value of 0 if x = y but the question says x!=y.
So, my question is: can we not choose (x - y) ^2 > 0 instead of choosing 2xy > 0? Because if we choose RHS of the inequality then, we arrive at the answer as not sufficient.
But if we choose it to be (x-y)^2 > 0 from RHS of the inequality equation, can we not deduce that (x -y )>0 and then take -y to the right side to make it x > y, thus making the answer sufficient.

I am sure I am violating some rule of inequality or absolute value but cannot put a finger on it. I wanted to know the reason why we cannot choose (x - y) ^ 2 >0, is it because even if we choose it, we cannot remove its squares, or I am completely off the mark and misunderstood this piece completely. 

The point here is that (x - y)^2 > 0 doesn't necessarily imply that x > y; it simply indicates that x ≠ y. This is because the inequality (x - y)^2 > 0 holds true regardless of whether x > y or x < y. It does not hold only if x ≠ y.

Hope this helps.

Bunuel Thanks for clarifying that. Your explanation helped me a lot to understand this. Now I am clear. (x - y) ^2 can mean both x > y and y> x. It makes a lot of sense. Appreciate your help. Cheers.