akhilsuhag wrote:Is |x²+y²| > |x²-y²| ?

(1) x > y

(2) x > 0

Alternate approach:
When both sides of an inequality are enclosed in absolute value symbols, we can SQUARE the inequality.

|x²+y²|² > |x²-y²|²

x� + y� + 2x²y² > x� + y� - 2x²y²

4x²y² > 0

(xy)² > 0.

In the resulting inequality, the left side will always be positive as long as xy≠0.
Question stem, rephrased:
Is xy ≠ 0?

Statement 1: x > y
Case 1: x=1 and y=0
In this case, xy=0, and the answer to the rephrased question stem is NO.
Case 2: x=2 and y=1
In this case, xy≠0, and the answer to the rephrased question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Cases 1 and 2 satisfy BOTH STATEMENTS.
Since the answer is NO in Case 1 but YES in Case 2, the two statements combined are INSUFFICIENT.

The correct answer is E.

A key takeaway from this problem is REMEMBERING that unless it's ruled out by information in either the question or the statements, the value of one or more variables in a DS question can be ZERO.

Great explanation from GMATGuruNY.

Thanks

Please suggest with what possible set of values we can evaluate such question

x,y ==> 2 1/2 0 -1/2 2

Can't we write it as below considering both sides positive.

x2+ y2 > x2 - y2

2y2 >0
y2>0
which means y > 0 always so we can't take y=0 & hence explanation from GMATGuruNY is doubtful.

Can't we write it as below considering both sides positive.

x2+ y2 > x2 - y2

2y2 >0
y2>0
which means y > 0 always so we can't take y=0 & hence explanation from GMATGuruNY is doubtful.