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.