inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2$
$x^5+4x^3+4x-2x^4-4x^2>0$
$x ( x^4 + 4x^2 + 4 +2x^3 -4x) >0 $
$x>0$ or ^ ... 
This solving method doesn't look right  

Break this up into cases.
Case 1: $x > 0$.
Multiplying both sides by $x$, we have $1 > \dfrac{2x^2}{x^2 + 2}$ (since $x$ is positive, the sign doesn't change), and so $x^2 + 2 > 2x^2$ (since $x^2 + 2$ is positive, the sign doesn't change).
Then $2 > x^2$. Now, just find the positive $x$ for which this is true.
Case 2: $x < 0$. 
Multiplying both sides by $x$, we have $1 < \dfrac{2x^2}{x^2 + 2}$ (since $x$ is negative, the sign changes), and so $x^2 + 2 < 2x^2$.
Then $2 < x^2$. Now, just find the negative $x$ for which this is true.
Edit: Also, to gain some intuition to see if whatever approach you choose works, you could plot the two functions, say by putting the following in the search bar in Chrome: "y = (2x/(x^2 + 2)) and y = 1/x". This won't give you an exact answer, but it'll let you know if you're barking up the right tree.
