Error in proof by contrapositive.

Question

Prove the assertion: if $x<-1$, then $x^2>1$

I am trying to prove this by contrapositive and so far I have

$x^2\ge1$ then $x\ge-1$

$x^2-1\ge0$

$(x-1)(x+1)\ge0$

$-1\le x \le 1$

which does not prove what I am trying to prove.

Answer 1

Why not prove it directly?

$$x<-1\implies \text{ (since}\;x\;\text{ is negative)} \;\; x^2>-x>1$$

the last inequality being the very first times $\;-1\;$ .