## Proof for $x\le -1 \implies x^3-x\le 0$ February 12

Here is my proof:Let $x\in \mathbb{R}$, assume $x\le -1$Then $x^2\ge 1$Then $x^3\le -1$Since $x\le -1$$x^3\le x$Then $x^3-x\le 0$Therefore $x\le -1 \implies x^3-x\le 0$Therefore $\forall$ $x\in \mathbb{R}$, $x\le -1 \implies x^3-x\le 0$I feel the pro