x^2+2x+A>0

△<0
4-4A<0
A>1

so x^2-2x+A>0 if A>1

Prove x^2+2x+A>0, this can be proved only when the value of A>0
Assume that is,
Let y=x^2+2x+A
y=(x+1)^2-(1-A)
Assume A gets minimum value 1
y=(x+1)^2>0
Assume A gets a much larger number, such as 50
y=(x+1)^2-(1-50)
y=(x+1)^2+49>0
If A is not >0 ,
Delta=(2)^2-4(1)(A)
=4(1-A)>0,The curve must have parts that is negative
Thus x^2+2x+A>0 cannot be proved