Triangle Inequality/Real Numbers/Proof 1

Theorem
Let $x, y \in \R$ be real numbers.

Let $\left\vert{x}\right\vert$ be the absolute value of $x$.

Then:
 * $\left\vert{x + y}\right\vert \le \left\vert{x}\right\vert + \left\vert{y}\right\vert$

Proof
Then by Order of Squares in Totally Ordered Ring:
 * $\left\vert{x + y}\right\vert \le \left\vert{x}\right\vert + \left\vert{y}\right\vert$