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  

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 $(x<-\sqrt2)$
