Proof by cases that $x \le |x|$? Proof by cases  that  $x \le |x|$?
May I have an example?
 A: $|x|$ is defined as $$|x|=\left\{
\begin{array}{rl}
x & x\geq 0 \\
-x & x<0 \\
\end{array}\right.$$
So when $x\geq 0$ than we have 
$|x| = x \geq x$ and this is known to be true. 
For $x<0$ we have 
$$x<0 < - (x)= |x|$$
A: Since
$$|x|=\begin{cases}-x&,\;x<0\\{}\\\;\;\;x&,\;x\ge 0\end{cases}$$
so for $\,x\in\Bbb R\,$ we have
$$x<0\Longrightarrow x<-x=|x|\Longrightarrow x<|x|\\x\ge 0\Longrightarrow x=x=|x|\le |x|\Longrightarrow x\le |x|$$
A: The answer depends on your definition of $|x|$.
For example, if your definition of $|x|=\max\{x,-x\}$ then there is no need for proof by cases, $x\leq\max\{x,-x\}$ by definition of $\max$.
Generally if you define $f(x)=\begin{cases} f_+(x) & x\geq 0\\ f_-(x) & x<0\end{cases}$ (in the case of $|x|$ we have $f_+(x)=x$ and $f_-(x)=-x$, but that's not to the point), then a proof by cases would be as follows:

Let $x$ be a real number. If $x\geq 0$ then $f(x)=f_+(x)$ [and derive the property wanted in that case]; otherwise $x<0$, and then $f(x)=f_-(x)$ [and derive the property wanted in this case as well].

