Studying this inequality:
$$x^3 -x -1 \ge 0 $$
Since I can't apply Ruffini's rule, I cannot recognize a method to study the function positivity. I could decompose it to:
$$ x \cdot (x+1) \cdot (x-1) -1 \ge 0$$
But the $-1$ is a problem, how do I go on?
You can study its derivative. Note that $f'(x)=3x^2-1$. This has roots at $x_1,x_2=\pm\dfrac 1{\sqrt 3}$, and one can see it is positive on $(x_1,\infty)$ and $(-\infty,x_2)$, and negative on $(x_2,x_1)$. Then, your function will be increasing on $(-\infty,x_2)\cup(x_1,\infty)$ and decreasing on $(x_2,x_1)$. Study its roots to conclude. As computing the roots is not easy, one can argue as follows: $x^3-x-1=0$ is equivalent to $x^3=x+1$. Because of the steep decrease of $x^3$ for negative values, we can argue there are no negative roots. On the other hand, $x^3\leq x+1$ for example for $x=0$, but $x^3\geq x+1$ for say $x=2$. So there must be a positive root in $[0,2]$.
Using Cardano's method on the depressed cubic one can find the positive root to be $$x=\left(\frac 1 2+\sqrt{\frac 1 4-\frac 1 {27}}\right)^{1/3}+\left(\frac 1 2-\sqrt{\frac 1 4-\frac 1 {27}}\right)^{1/3}$$
If you consider the function $f(x)=x^3-x-1$, you have $$ \lim_{x\to-\infty}f(x)=-\infty,\qquad \lim_{x\to\infty}f(x)=\infty $$ so you know there's at least a point where the function $f$ is zero. Consider now $$ f'(x)=3x^2-1 $$ that is zero at $\alpha=-1/\sqrt{3}$ and $\beta=1/\sqrt{3}$. This tells us that there is a relative minimum at $\alpha$ and a relative maximum at $\beta$. Now $$ f(\alpha)=-\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}}-1= \frac{2}{3\sqrt{3}}-1<0; $$ Similarly, $$ f(\beta)=\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}-1=-\frac{2}{3\sqrt{3}}-1<0. $$ Since the relative maximum is negative, this means that the equation $x^3-x-1=0$ has only one root $x_0$. Its location can be further delimited by observing that $f(1.3)=-0.103<0$ and $f(1.4)=0.344>0$, so that $1.3<x_0<1.4$. With cleverer numerical methods you can better approximate the root. The conclusion is that $x^3-x-1>0$ for $x>x_0$.
A different way to get the same conclusion is to note that the equation $x^3-x-1=0$ is a depressed cubic (no second degree term), so one can look at its discriminant $$ \frac{p^3}{27}+\frac{q^2}{4}=-\frac{1}{27}+\frac{1}{4}>0 $$ which directly says that the equation has only one real root. Write the equation as $x^3+px+q=0$ for finding $p$ and $q$.
Let's study the number of real roots of the polynomial $x^3-x-1$. We look at its derivative, $3x^2-1$, and conclude that it has local maximum at $x_M=-1/\sqrt 3$ and a local minimum at $x_m=1/\sqrt 3$. The values of the polynomial at these points are $\frac {2}{3\sqrt 3}-1 $ and $-\frac {2}{3\sqrt 3}-1$, respectvely. Clearly, if the value in local maximum is negative, then there's only one real root (remember that our polynomial goes to $+\infty$ as $x\to\infty$). We arrive to the conclusion that $x^3-x-1$ has only one real root $x_\ast$, which can be found either analytically or numerically. On $(x_\ast,\infty)$ our polynomial is positive, on $(-\infty,x_\ast)$ it's negative.
