Manipulating the triangle inequality to compare two sides

Question

With $\mathbf x$ and $\mathbf y$ two vectors, the triangle inequality can be written as:

$$ |\mathbf x|-|\mathbf y| \le |\mathbf x + \mathbf y|\le |\mathbf x|+|\mathbf y| \tag{A} $$

which can be broken down into a system of inequalities (labelled as (1) and (2)):

\begin{equation} \begin{cases} |\mathbf x|-|\mathbf y| \le |\mathbf x + \mathbf y| \\ |\mathbf x + \mathbf y|\le |\mathbf x|+|\mathbf y| \end{cases} \end{equation}

then, with the assumption of $|\mathbf y|>0$, (1)-(2) yields:

$$ |\mathbf x| \le |\mathbf x + \mathbf y| \tag{B} $$

• But is this in fact true? Is (B) a direct consequence of (A) when $|\mathbf y|>0$? or is (B) is invalid because (1)-(2) is not allowed as it is not obvious if the subtraction preserves the inequality?

This question arose during my attempt to this problem, where I build triangles from 3 points on a line in order to try and show which point is the last one (to no avail so far).

Answer 1

No, we cannot subtract inequalities in the following way: $$ a \leq b\\ c\leq d\\ \implies a-c\leq b-d $$ For example, take $a=3,b=4,c=1,d=3$ to see that it's wrong. I think this is what you did to obtain what you have. So $$|\mathbf x|\leq |\mathbf x+\mathbf y| $$ is wrong. A counterexample in $\mathbb{R}^2$: $\mathbf x=(1,0),\mathbf y=(-1,0)$.

Answer 2

(B) does not follow from (1)-(2). For a very simple counter-example, work in the 1-dimensional space of real numbers and take x=3, y=-2.