Solve the following inequalities over the set of real numbers:

a) $\sqrt{x^{2}-x-2} \leq 2 x$,

b) $\sqrt{x^{2}-x-2} \geq 2 x$.
Under the square root, there can only be a non-negative number. Therefore, we must constrain that $x^{2}-x-2 \geq 0$. The function $f(x)=x^{2}-x-2$ represents an upward-opening parabola, which intersects the x-axis at two points: -1 and 2. Therefore, the values of the function are negative on the interval $(-1 ; 2)$, and positive on the intervals $(-\infty ;-1)$ and $(2 ; \infty)$. Thus, the constraint is: $x \leq-1$ or $x \geq 2$.

If $x \leq-1$, then $\sqrt{x^{2}-x-2}$ is non-negative, and $2x$ is negative, so the inequality b) holds, while a) does not. If $x \geq 2$, then $\sqrt{x^{2}-x-2}<\sqrt{x^{2}}=x<2x$, so the inequality a) holds, while b) does not.

In summary: the solution to inequality a) is $x \geq 2$, and the solution to inequality b) is $x \leq-1$. In neither a) nor b) can equality hold.