<s>[INST] Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\frac{x}{2}<\frac{2}{x}\Leftrightarrow\frac{(x-2)(x+2)}{2x}<0\Leftrightarrow x<-2.$
Since both of these inequalities have to be satisfied simultaneously, one can combine them to get $x<-2.$ Correct answer is $x\in(\sqrt{2},2)$
 [/INST] There are some good answers up already but let me address specifically what's wrong with your attempted solution.
1) $1/x < x/2$ is not equivalent to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x} < 0$. The inequality is going the wrong way. Subtracting $x/2$ on the left side gives
$$\frac{1}{x} - \frac{x}{2} = \frac{2 - x^2}{2x} = \frac{(\sqrt{2}-x)(\sqrt{2}+x)}{2x},$$
so this is what is $<0$.
2) Meanwhile this inequality is not equivalent to $x<-\sqrt{2}$. The numerator is positive when $x$ is between $-\sqrt{2}$ and $\sqrt{2}$ and negative otherwise. The denominator changes sign with $x$. Thus the fraction is pos/neg = neg when $-\sqrt{2}<x<0$, and neg/pos = neg when $x>\sqrt{2}$, so overall it is negative when either $-\sqrt{2}<x<0$ or $x>\sqrt{2}$.
3) For $x/2< 2/x$, this time it is equivalent to $(x-2)(x+2)/2x<0$, as you wrote. (It's different from the last one because the $x$ is in the numerator on the left side.) But again, this is not equivalent to $x<-2$. Now the numerator is positive when $x<-2$ or $x>2$ and negative between $-2$ and $2$, while again the denominator changes sign with $x$. Thus the whole fraction is pos/neg = neg when $x<-2$, and it is neg/pos = neg when $0<x<2$. So this inequality yields $x<-2$ or $0<x<2$.
</s>