Can I proof this statement "Suppose $x \in \mathbb{R}$ If $x^3-x>0$ then $x>-1$" like this?

Question

Proof by contrapositive

We want to prove "If $x \leqslant-1$ then $x^3-x\leqslant0$"

Suppose $x\leqslant-1$ Then $x^3\leqslant-1^3=-1$ Thus $x^3-x\leqslant0$ Therefore $x^3-x\leqslant0$ #

Answer 1

You see $y=x^3-x=x(x-1)(x+1)$ has roots $x=-1,0,1$. If $x<-1$ then $y$ will never touch zero again. Test by $x=-2$ and you have $y=-6$ so all $y$ are negative for $x<-1$. In the same way, if $x>1$ then $y$ is positive since $y=6$ for $x=2$.

We can add that all polynomials are continuous so there are no discontinuous jumps over zero.

Answer 2

If $x \leqslant -1$ then $x^{3}-x=(-x) \cdot (1-x^{2})$ which is the product if a positive number and a non-positive number. Hence $x^{3}-x\leqslant 0$.

Answer 3

$x\le -1\implies x^3\le -1\implies x^3-x=x^3+|x|\le -1+|x|=??$

There are many ways to do this correctly. Let $x=-r-1.$ Then $$x^3-x=x(x^2-1)=x((-r-1)^2-1)=x(r^2+2r)=x\cdot (r)(r+2).$$ Now we have $$x\le -1\implies (x<0\, \land \, r\ge 0)\implies$$ $$ (x<0\,\land r\ge 0\,\land \, r+2 \ge 0+2>0)\implies$$ $$ (x<0\,\land (r)\,(r+2)\ge 0)\implies$$ $$ x\cdot (r)(r+2)\le 0.$$