Alternate Triangle Inequality: Instead of $\Big|\,|x| - |y|\,\Big| \leq |x-y|$ can we use $|x| - |y| \leq |x-y|$?

Question

Does the Alternate Triangle Inequality hold if we do not use the outer mod?

That is, instead of $$\Big|\,|x| - |y|\,\Big| \leq |x-y|$$

can we use the following? $$|x| - |y| \leq |x-y|$$

Thanks in advance

Answer 1

$$\begin{align} T_\text{w}&:& |x|-|y| \leq |x-y|\quad&\forall x,y\in S \end{align}$$ is equivalent to $$\begin{align} T_\text{s}&:& \bigl||x|-|y|\bigr| \leq |x-y|\quad&\forall x,y\in S. \end{align}$$

Proof: from $T_\text{s}$ follows $T_\text{w}$ immediately because $a\leq|a|$ holds always. In the other direction, we need to make a case distinction:

• If $|x|\geq|y|$, then $\bigl||x|-|y|\bigr| = |x|-|y|$ and thus $T_\text{s}$ is $T_\text{w}$.
• If $|x|<|y|$, then $$ \bigl||x|-|y|\bigr| = |y|-|x| =: |\tilde{x}|-|\tilde{y}|, $$ which, since $T_\text{w}$ quantifies over all $x,y\in S$, is $$ \leq|\tilde{x}-\tilde{y}| = |y-x| = |x-y|, $$ which was to be shown.

Answer 2

$|x-y| \ge |x|-|y|$ will always be true and it will be trivial.

Answer 3

just check that $$ |(x - y) + y| \leqslant |x - y| + |y| \implies |x| - |y| \leqslant |x - y| $$ $\color{red}{\text{The inequality }} \color{blue}{ ||x| - |y|| \leqslant |x - y| \text{ is the general case } }$