find all real values of x such that (x+1) / (x^2+3) < 2/x?

Question

i started with: (x-1) / (x^2+3) - 2/x < 0
then found common denominator, and after simplifying i got a quadratic on top and incomplete cubic on the bottom... it doesn't seem right though...and i can't solve it any further... please help.

Answer 1

(x+1) / (x^2+3) < 2/x
Since x^2 + 3 > 0 we can multiply throughout by it
(x+1) < 2*(x^2 + 3)/x

Case 1: If x > 0
x*(x+1) < 2*(x^2 + 3)
x^2 - x + 6 > 0
This is +ve for all x > 0

Case 2: If x < 0
x*(x+1) > 2*(x^2 + 3)
x^2 - x + 6< 0
which is never the case

ANS: x > 0

Answer 2

(x-1)/(x^2+3)-2/x<0
(x^2-x-2x^2-6)/(x(x^2+2)
The sign of the denominator is the sign of x as x^2+3>0
-(x^2+x+6)= always negative
so the solution is x>0