Solution for the inequality $x \lt \frac 1 x$ I am unable to find correct solution for the inequality:
$x \lt \frac 1 x, x \in \Bbb R, x \ne 0$.
My solution first takes $x > 0$, then we have:
$x < \frac 1x$ (multiply both sides by $x$)
$\iff  x^2 < 1$ (taking square root)
$\iff  x < 1 \lor x < -1$
$\iff  0 < x < 1$ since $x > 0$.
Then similarly, for $x < 0$,
$x < \frac 1x$ (multiply both sides by $x< 0$)
$\iff  x^2 > 1$ (taking square root)
$\iff  x > 1 \lor x > -1$
$\iff  -1 < x < 0 $ since $x < 0$
But this does not agree with the solution $(x < -1) \lor (0 < x < 1)$
Which part of my solution is incorrect above?

$$x<\frac { 1 }{ x } \\ x-\frac { 1 }{ x } <0\\ \frac { { x }^{ 2 }-1 }{ x } <0\\ \frac { x\left( x-1 \right) \left( x+1 \right)  }{ { x }^{ 2 } } <0\\ x\left( x-1 \right) \left( x+1 \right) <0\\ x\in \left( -\infty ;-1 \right) \cup \left( 0;1 \right) $$
