Reverse Triangle Inequality/Real and Complex Fields/Proof 2

Proof
From proof $2$ of corollary $1$ to this result, which is derived independently:
 * $\size {x - y} \ge \size x - \size y$

There are two cases:

$(1): \quad \size x \ge \size y$

We have :
 * $\size {\size x - \size y} = \size x - \size y$

and the proof is finished.

$(2): \quad \size y \ge \size x$

We have:
 * $\size {y - x} \ge \size y - \size x = \size {\size y - \size x}$

But:
 * $\size {y - x} = \size {x - y}$

and:
 * $\size {\size y - \size x} = \size {\size x - \size y}$

From this we have:
 * $-\size {\size x - \size y} \ge -\size {x - y}$

Since, by Negative of Absolute Value, we have that:
 * $\size x - \size y \ge -\size {\size x - \size y}$

it follows that:
 * $-\size {x - y} \le \size x - \size y \le \size {x - y}$

The result follows.