How to prove $|y|≤|x|+|z|$ if $ x≤y≤z$?

Question

How to prove $|y|≤|x|+|z|$ if $ x≤y≤z$?

I have tried to treat it as a triangular inequality but it came to nothing. Can someone think of how to try it?

Answer 1

If $y \ge 0$ then $$ |y| = y \le z = |z| \le |x| + |z| $$ and if $y < 0$ then $$ |y| = -y \le -x = |x| \le |x| + |z| $$

Alternatively $$ |y| = \max(-y, y) \le \max(-x, z) \le \max(|x|, |z|) \le |x| + |z| \, . $$

Answer 2

If $y \geq 0$, then $z \geq y$ is a larger (or equal) positive number and hence $|y| \leq |z|$. Since $|x|$ is necessarily a positive number, we can conclude that $|y| \leq |z| + |x|$.

If $y < 0$, then $x \leq y$ is a smaller (or equal) negative number and hence $|y| \leq |x|$. Since $|z|$ is necessarily a postive number, we can conclude that $|y| \leq |z| + |x|$.