proof with inequalities simple question

Question

How do you prove:

Suppose $x$ is a real number. if $x^3-x>0$ then $x>-1$

It seems really easy to do the contrapositive here i think but dont now how to word it.

So suppose $x \le 1$ then $x^3 \le x$ for all x? would this be ok Thanks

Answer 1

$$ x(x-1)(x+1)>0\implies x\in(-1,0)\cup (1,\infty)\implies x>-1$$

Answer 2

The contrapositive of a statement "$P \Rightarrow Q$" is "$\neg Q \Rightarrow \neg P$."

The negation of "$x > -1$" would not be "$x \leq 1$", it would be "$x$ is NOT greater than $-1$" - equivalently, "$x \not > -1$" or "$x \leq -1$".

You negate what is being said, not the actual signage in the statement.