3.1.3 Specific Transformations

For the first inequality, the value $x=0$ is not in the domain, hence this value is excluded. For $x>0$, taking the reciprocal while inverting the comparator is allowed and results in $x\le 3$. Together with the condition above the solution interval is $L=]0;3]$. For $x<0$ the reciprocal rule cannot be applied. However, it can be seen, even without any rule, that none of the values $x<0$ can be a solution, since then $\frac{1}{x}$ is negative as well and not greater than or equal to $\frac{1}{3}$.
The domain of the second inequality is $]0;\infty [$, since only for these values of $x$ taking the square root is defined and only for $x\ne 0$ the denominators are non-zero. On the domain, taking the reciprocal while inverting the comparator is allowed and results in $x>\sqrt{x}$. Since $\sqrt{x}>0$, the inequality can be divided by $\sqrt{x}$ resulting in $\sqrt{x}>1$. This inequality has the solution set $L=]1;\infty [$ which occurs also in the domain.