Proving $\forall x, -|x| \le x \le |x|$ 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 $-|x|  \leq x\leq |x|$.
 A: Hint :  use the definition  $$\lvert x \rvert = \cases {{\ x, \text {if } x\ge 0\\ {-x}, \text  {if } x \lt 0}}$$
A: 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.
A: 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:

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

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