Solve the inequality 1/x > x I am trying to solve this inequality but I am repeatedly getting an incorrect answer. So I first set the restriction $x$ cannot be $0$ and then split it into two cases. One where $x > 0$ and another where $x < 0$.
When $x$ is above $0$, I  get $1 > x^2\implies -1 < x$ and $ x < 1$. But $x < -1$ is not valid in this case as it has been declared that $x$ is above $0$ in this case so you get $0 < x < 1$ which is correct.
The other case is when $x$ is below $0$ so I must flip the sign because I am multiplying by $x$ when it is negative. I get $1 < x^2$, $1 < x$,$-1 < x$. This is incorrect because the domain must be that $x < -1$, I understand why my answer is wrong logically but I do not know where I made the mistake in my work.
Here is picture of my work:
Image
 A: Another way
$$\frac1x>x \iff \frac1x-x>0\iff \frac{1-x^2}{x}>0$$
then

*

*for $x>0$ we need

$$1-x^2>0 \iff x^2<1 \iff -1<x<1$$
that is
$$0< x<1$$

*

*for $x<0$ we need

$$1-x^2<0 \iff x^2>1 \iff x<-1 \lor x>1$$
that is
$$x<-1$$
and then
$$x\in (-\infty,-1)\cup(0, 1)$$
Here is a plot to visualize the solution


As noticed in the comments your mistake is that for $x<0$ isn’t true that
$$1<x^2\implies 1<x \lor -1<x$$
since $\sqrt {x^2}=|x|$ which leads to
$$1<x^2\implies 1<|x|\implies 1<-x\implies x<-1$$
A: For $\;x>0\;$:
$$\frac1x>x\implies x^2<1\implies |x|<1\implies 0<x<1$$
For $\;x<0\;$ :
$$\frac1x>x\implies x^2>1\implies|x|>1\implies x<-1$$
A: Since you are given that $\frac{1}{x}>x$ you know that $x\ne0$ since you cannot divide by $0$.
Since multiplication of an inequality by a negative quantity reverses the direction of the inequality, one might instead multiply by $x^2$ rather than by $x$. This gives
\begin{eqnarray} x&>&x^3\\
x-x^3&>&0\\
x(1-x^2)&>&0\\
x(1-x)(1+x)&>&0
\end{eqnarray}
The expression will be non-zero in each of the four intervals $(-\infty,-1),(-1,0),(0,1), (1,\infty)$ but will be positive in only the intervals $(-\infty,-1)$ and $(0,1)$. So the solution of the inequality is
$$ x<-1\quad\text{or}\quad0<x<1 $$
A: Given that
$0 < x < \dfrac{1}{x}, \;\; (1)$
we may multiply through by $x$ and find
$0 < x^2 < 1; \;\; (2)$
this in turn implies
$x < 1,  \;\; (3)$
since
$x \ge 1 \;\; (4)$
immediately yields
$x^2 \ge 1, \;\; (5)$
in contradiction to (2).  Combining (1) and (3) yields
$0 < x < 1, \;\; (6)$
and dividing this through by $x > 0$ we obtain
$0 < 1 < \dfrac{1}{x}; \;\; (7)$
(6) and (7) together beget
$0 < x < \dfrac{1}{x}. \;\; (8)$
We have now shown that every $x$ satisfying (1) lies in the interval $(0, 1)$; that is,  $(0, 1)$ contains all positive solutions to
$x < \dfrac{1}{x}, \;\; (9)$
and that every $x \in (0, 1)$ satisfies (9).
Now suppose (9) holds with
$x < 0; \;\; (10)$
then since
$x < 0 \Longleftrightarrow \dfrac{1}{x} < 0, \;\; (11)$
we can assume that
$x < \dfrac{1}{x} < 0. \;\; (12)$
We multiply (12) through by $x < 0$, and obtain
$x^2 > 1 > 0, \;\; (13)$
which together with $x < 0$ forces
$x < -1 < 0; \;\;(14)$
thus every solution of (12) obeys (14).  Now if $x$ satisfies (14), dividing through by $-x$ gives
$-1 < \dfrac{1}{x} < 0; \;\; (15)$
then combining (14) and (15) we have
$x < \dfrac{1}{x} < 0, \;\; (16)$
in agreenent with (12), and we are done.
It will be noted that the complete set of solutions $X$ to
$x < \dfrac{1}{x} \;\; (17)$
is the union of two disjoint open intervals:
$X = (-\infty, -1) \cup (0, 1); \;\; (18)$
furthermore, the map
$\theta(y) = -\dfrac{1}{y}   \;\; (19)$
is easily seen to carry $(0, 1)$ onto $(-\infty, -1)$ and vice-versa; it is also obviously injective.  We apply $\theta$ to (17):
$\theta(x) = -\dfrac{1}{x},  \;\; (20)$
$\theta \left(\dfrac{1}{x} \right) = -x;  \;\; (21)$
then (17) may be written
$-\theta \left(\dfrac{1}{x} \right) < -\theta(x), \;\; (22)$
and multiplying this by $-1$ we obtain
$\theta(x) < \theta \left(\dfrac{1}{x} \right), \;\; (23)$
which shows that the relation expressed by (17) is invariant under the action of $\theta$.
