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  
 A: 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.
A: Since $x^2+2>0$,
$$\frac{1}{x}>\frac{2x}{x^2+2}\iff \frac{x^2+2}{x}=x+\frac{2}{x}>2x\iff x<\frac{2}{x}$$  
$$\iff \begin{cases}\begin{cases}x>0\\ x^2<2\end{cases}\\\ \ \ \text{or}\\ \begin{cases}x<0\\ x^2>2\end{cases}\end{cases}\iff \begin{cases}\begin{cases}x>0\\ -\sqrt{2}<x<\sqrt{2}\end{cases}\\\ \ \ \text{or}\\ \begin{cases}x<0\\ x\in(-\infty,-\sqrt{2})\cup(\sqrt{2},+\infty)\end{cases}\end{cases}$$   
$$\iff \begin{cases}0<x<\sqrt{2}\\\ \ \ \text{or}\\ x\in(-\infty,-\sqrt{2})\end{cases}$$
A: As both sides of the inequation have the same sign, it is equivalent to:
$$\frac{x^2+2}{x}=x + \frac2x> 2x\iff x<\frac2 x\iff\begin{cases}x^2<2&\text{if}\enspace x>0,\\x^2>2&\text{if}\enspace x<0\end{cases}$$
Thus the solutions are $\,(0< x<\sqrt 2)\,$ or $\,(x<-\sqrt 2)$.
A: We need $$\dfrac1x-\dfrac{2x}{x^2+2}>0$$
$$\iff\dfrac{x^2+2-2x^2}{x(x^2+2)}>0$$
As $x^2+2>0$ for real $x,$ multiplying both sides by $x^2(x^2+2)$
$$\iff x(2-x^2)>0\iff\{x-(-\sqrt2)\}\cdot x\cdot (x-\sqrt2)<0$$
Clearly, we need to check for $(-\infty,-\sqrt2);(-\sqrt2,\sqrt2);(\sqrt2,\infty)$
As the product of three terms is negative, we need exactly one $(0<x<\sqrt2)$ or all three to be negative $(x<-\sqrt2)$
A: In response to @Bernard's answer, I offer the graph of both sides of the inequality.

I respect the member who DVs this as 'not an answer,' but, of course, not every answer here is a complete answer. A hint has value, as does a method of confirming one's algebraic solution. 
The graph's intercepts, where we set the inequality to an equality, does a good job offering the two points of interest, where the curves cross. $\,(x+/-\sqrt 2)$
Again, not a full solution, but an aid to verification. 
