Prove that $0 \leq \frac{x+|x|}{2} \leq |x|$

Question

$\frac{x+|x|}{2}$ is superior or equal to $0$ but inferior or equal to $|x|$ where the $x$ is a real number.

I must prove this by the method of proof by cases.

I have no idea one how to begin this. Can somebody give me a hint?

Answer 1

By cases, it is immediate (one case leads to $|x|$ and the other to $0$).

A cleaner approach would be:

$$\frac{x+|x|}{2} \leq \frac{|x|+|x|}{2} \leq \frac{2|x|}{2} \leq |x|.$$