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 $|x - y| \ge |x| - |y|$:

From Triangle Inequality we see $|x + y| - |y| \le |x|$.

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


 * $|z| - |y| \le |z - y|$

Renaming variables as appropriate gives:
 * $|x - y| \ge |x| - |y|$


 * Next we show that $|x - y| \ge ||x| - |y||$:

From $|x - y| \ge |x| - |y|$ (proved above), we have:

If $x \ge y,$
 * $||x| - |y||=|x| - |y|,$ and we're done.

If $y \ge x, $
 * $|y - x| \ge |y| - |x| = ||y| - |x||$.

But $|y-x|=|x-y|$ and $||y|-|x||=||x|-|y||$

Note from this we have:
 * $-||x|-|y|| \ge -|x-y|$

Since, by Negative of Absolute Value, we have that $|x|-|y| \geq -||x|-|y||$, it follows that:
 * $-|x-y| \le |x| - |y| \le |x-y|$