Question

If x > -x^2 and y > 1, then prove that xy < 0 for all possible values of x and y.

Answer 1

The question is wrong. I will prove it by just 1 counter example.
Let x = 1.
Then -x^2 = -1
so x > -x^2 is satisfied since 1 > -1.
So x= 1 and y > 1 and xy > 1 which contradicts xy < 0

Maybe the question was x < -x^2 (1)
and y > 1
Consider 2 cases: x < 0 and x > 0
Case I: x >0
Divide inequality (1) on both sides by x to get
1 < -x which is impossible since x > 0 and therefore -x < 0
So this case is not valid
Case II: x < 0
Divide inequality (1) on both sides by x to get:
1 > -x (2)
[The inequality is flipped from < to > since we divided by negative number x]
Now since x < 0, -x > 0 so a solution to (2) is -1 < x < 0.
Since we have a valid solution with x < 0 only, and y >1, it follows that xy will be negative.