Reverse Triangle Inequality/Real and Complex Fields/Corollary 1

Theorem
Whether $x$ and $y$ are in $\R$ or $\C$, 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 $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$:

From Triangle Inequality we see $\left\vert{x + y}\right\vert - \left\vert{y}\right\vert \le \left\vert{x}\right\vert$.

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:

Also:
 * $\left\vert{x - y}\right\vert = \left\vert{y - x}\right\vert \ge \left\vert{y}\right\vert - \left\vert{x}\right\vert = - \left({\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right)$

Thus:
 * $- \left({\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right) \le \left\vert{x - y}\right\vert \le \left({\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right)$

from the corollary to Negative of Absolute Value.