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 $x\in(\sqrt{2},2)$
 A: First, notice that $1/x < 2/x$ implies $1/x$ is positive, hence $x$ is positive. Now you know you can multiply the inequalities by $x$ without reversing the inequality signs.  So you get $2<x^2$ and $x^2<4$.  And so on...
A: If $x>0$, multiply by $x$ to get
$$1<\frac{x^2}{2}<2\iff 2<x^2 <4\implies \sqrt 2<x<2$$
If $x< 0$ multiply by $x$ and 
$$1>\frac{x^2}{2}>2\iff 2>x^2>4\implies \text{false}$$
A: Hint :$$\frac{1}{x}<\frac{x}{2} \to x\in (-\sqrt 2,0) \cup(\sqrt 2,+\infty)\tag{1}$$
$$\frac{x}{2}<\frac{2}{x}\to x\in(-\infty,-2)\cup (0,2) \tag{2}$$ then find $(1) \cap (2)$
A: $\dfrac1x<\dfrac2x$ is only possible for positive $x$. 
Then we may multiply by $2x$ and take the square roots,
$$\sqrt2<x<2.$$
A: 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 $0<x<2$.
