Negative of Absolute Value

Theorem
Let $$x \in \mathbb{R}$$ be a real number.

Let $$\left|{x}\right|$$ be the absolute value of $$x$$.

Then $$- \left|{x}\right| \le x \le \left|{x}\right|$$.

Proof
Either $$x \ge 0$$ or $$x < 0$$.


 * If $$x \ge 0$$, then $$- \left|{x}\right| \le 0 \le x = \left|{x}\right|$$.


 * If $$x < 0$$, then $$- \left|{x}\right| = x < 0 < \left|{x}\right|$$.