<s>[INST] 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  
 [/INST] 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 $\,(x<-\sqrt 2)$.
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