Triangle Inequality proof with three variables

Question

How do I prove $$|x + y+ z| \le |x|+|y|+|z|$$ without using the triangle inequality itself?

Are cases the best way to go? Or using properties of absolute? very stuck

Answer 1

Because $$|x|+|y|+|z|=\sqrt{(|x|+|y|+|z|)^2}=$$ $$=\sqrt{x^2+y^2+z^2+2|xy|+2|xz|+2|yz|}\geq$$ $$\geq\sqrt{x^2+y^2+z^2+2xy+2xz+2yz}=$$ $$=\sqrt{(x+y+z)^2}=|x+y+z|.$$

Answer 2

We can make use of two properties:

$\mathbf{1}.$ For some constant $k\geq0$: $$|x|\leq k\iff-k\leq-x\leq k$$

$\mathbf{2.}$ For $x\in \mathbb{R}$, $$-|x|\leq x\leq |x|$$

The proof for the triangle inequality in three variables is as follows:

For $x,y,z\in \mathbb{R}$

Property $\mathbf{2.} \implies \begin{cases} -|x|\leq x\leq|x| \\ -|y|\leq y\leq |y| \\ -|z|\leq z \leq |z| \end{cases} $

Summing each inequality yields:

$$-(|x|+|y|+|z|)\le x+y+z\leq |x|+|y|+|z|$$

Property $\mathbf{1.}\implies |x+y+z|\leq|x|+|y|+|z|$