Proving $\forall x, -|x| \le x \le |x|$

Question

I'm trying to prove the following inequality $\forall x, -|x| \leq x \leq |x|$.

Let us try to proceed by reasoning by cases:

Case 1: $x \geq 0$

When $x \geq 0$, we know that $-x \leq x$. Also, $x \leq x$. Combining them, we get $-x \leq x \leq x$. Since $x \geq 0$, it can be re-written as $-|x| \leq x \leq |x|$

Case 2: $x < 0$

This is the case where I'm stuck. I know here that $-x > x$. How to proceed from here to prove $-|x| \leq x\leq |x|$.

Answer 1

If $x < 0$, then $-x = |x|$, or equivalently, $x = -|x|$. Putting these together with $x < -x$, we get $$-|x| = x < -x = |x|$$ which immediately implies case 2.

Answer 2

Choose any $x \in \mathbb R$. Then we know that:

$$|x| = -x \qquad\text{ or }\qquad |x| = x$$

Equivalently, we know that:

$$x = -|x| \qquad\text{ or }\qquad x = |x|$$

But since $-|x| = -|x| \leq |x|$ is true and $-|x| \leq |x| = |x|$ is true, we conclude that:

$$-|x| \leq x \leq |x|$$

Answer 3

Case 2: $x < 0$

In this case $\,-x \gt 0\,$, and applying the inequality established for Case 1 to $-x$ it follows that:

$$ -|-x| <= -x <= |-x| $$

But $\,|-x|=|x|\,$ by definition, then multiplying by $\,-1\,$ (which reverses the inequalities) gives:

$$ |x| \ge x \ge -|x| $$

The latter is precisely the inequality that needed to be proved, which concludes the proof in Case 2.

Answer 4

$$0\leq \big(|x|-x\big)^2=|x|^2-2x|x|+x^2=2|x|^2-2x|x|=2|x|(|x|-x)\Longrightarrow |x|-x\geq 0$$$$\Longrightarrow |x|\geq x$$ since $|x|^2=x^2$ and $|x|\geq 0$ for all $x$. Similarly, $$0\leq \big(|x|+x\big)^2=|x|^2+2x|x|+x^2=2|x|^2+2x|x|=2|x|(|x|+x)\Longrightarrow |x|+x\geq 0$$$$\Longrightarrow x\geq -|x|.$$ Thus we have $\,-|x|\leq x\leq |x|$

Answer 5

Hint : use the definition $$\lvert x \rvert = \cases {{\ x, \text {if } x\ge 0\\ {-x}, \text {if } x \lt 0}}$$

Answer 6

In Case $1$ you got: $$-|x|\le x \le |x|, \ \ \ x\ge 0.$$ Now multiply the inequalities by $(-1)$ to get Case $2$: $$|x|\ge -x\ge -|x|, \ \ \ -x\le 0 \stackrel{-x=y} \Rightarrow |-y|\ge y \ge -|-y|, \ \ \ y\le 0 \Rightarrow |y|\ge y \ge -|y|, \ \ \ y\le 0.$$

Answer 7

to do it your way:

Case 2:

$x < 0$. So $ x < 0 < -x$ and $x=-|x|$ and $|x| = - x$.

So $-|x| = x \le x < -x = |x|$.

.....

But I think it is more straightforward to just note:

By definition $|x| \ge 0$ and so $-|x| \le 0$ so $-|x| \le 0 \le |x|$.

And also by definition either $x = |x|$ or $x = -|x|$.

If $x = |x|$ (which happens whenever $x \ge 0$) then

$-|x| \le x = |x|$.

If $x = -|x|$ (which happens whenever $x \le 0$) then

$-|x| = x \le |x|$.

Either way $-|x| \le x \le |x|$.

That's all.