Reverse Triangle Inequality/Real and Complex Fields/Corollary 1

Theorem
Whether $x$ and $y$ are in $\R$ or $\C$, or in fact in any normed vector space, the following hold:


 * $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$


 * $\left\vert{x - y}\right\vert \ge \left\vert{\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right\vert$

Proof

 * First we show that $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$:

By the triangle inequality, $\left\vert{x + y}\right\vert - \left\vert{y}\right\vert \le \left\vert{x}\right\vert$.

Substitute $z = x + y \implies x = z - y$ and so:


 * $\left\vert{z}\right\vert - \left\vert{y}\right\vert \le \left\vert{z - y}\right\vert$

Renaming variables as appropriate gives:
 * $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$


 * Next we show that $\left\vert{x - y}\right\vert \ge \left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert$:

From $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$ (proved above), we have:

If $\left\vert{x}\right\vert \ge \left\vert{y}\right\vert$:
 * $\left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert = \left\vert{x}\right\vert - \left\vert{y}\right\vert$, and we're done.

If $\left\vert{y}\right\vert \ge \left\vert{x}\right\vert$:
 * $\left\vert{y - x}\right\vert \ge \left\vert{y}\right\vert - \left\vert{x}\right\vert = \left\vert{ \left\vert{y}\right\vert - \left\vert{x}\right\vert }\right\vert$.

But $\left\vert{y - x}\right\vert = \left\vert{x - y}\right\vert$ and $\left\vert{ \left\vert{y}\right\vert - \left\vert{x}\right\vert }\right\vert = \left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert$.

Note from this we have:
 * $-\left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert \ge -\left\vert{x - y}\right\vert$

Since, by Negative of Absolute Value, we have that $\left\vert{x}\right\vert - \left\vert{y}\right\vert \ge -\left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert$, it follows that:
 * $-\left\vert{x - y}\right\vert \le \left\vert{x}\right\vert - \left\vert{y}\right\vert \le \left\vert{x - y}\right\vert$

The result follows.