Question stem, rephrased:VJesus12 wrote:Is xy<0?
1) |x+y| < |x| + |y|
2) |x| - |y| < | x - y |
Do x and y have different signs?
Statement 1: |x+y| < |x| + |y|
Since an absolute value cannot be negative, both sides here are NONNEGATIVE, allowing us to SQUARE the inequality:
(|x+y|)² < (|x| + |y|)²
x² + y² + 2xy < x² + y² + 2|x||y|
2xy < 2|x||y|
xy < |x||y|.
The resulting inequality is valid only if x and y have DIFFERENT SIGNS.
Thus, the answer to the question stem is YES.
SUFFICIENT.
Statement 2: |x| - |y| < |x-y|
Here, the left side will be positive if |x|>|y| and negative if |x|<|y|.
Since the sign of the left side is unknown, it might be safer to test cases.
Case 1: x=1 and y=-2, with the result that |x|-|y|=-1 and |x-y|=3
In this case, x and y have different signs, so the answer to the question stem is YES.
Case 2: x=-1 and y=-2, with the result that |x|-|y|=-1 and |x-y|=1
In this case, x and y have the same sign, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.
The correct answer is A.
