Triangle Inequality/Real Numbers/Proof 4

Proof
We do a case analysis.
 * $x\ge0, y\ge 0\Rightarrow \left\vert{x + y}\right\vert = x+y=\left\vert{x}\right\vert + \left\vert{y}\right\vert$


 * $x\le0, y\le 0\Rightarrow \left\vert{x + y}\right\vert = -x-y=\left\vert{x}\right\vert + \left\vert{y}\right\vert$


 * $x\ge0, y\le 0$. We have $\left\vert{x}\right\vert = x$ and $\left\vert{y}\right\vert = -y$.
 * In this case we show


 * $\displaystyle\left\vert{x + y}\right\vert \le \max\left(\left\vert{x}\right\vert, \left\vert{y}\right\vert\right).$


 * If $\left\vert{x}\right\vert \le \left\vert{y}\right\vert$, then


 * If $\left\vert{x}\right\vert \ge \left\vert{y}\right\vert$, then

We have $\max(a,b)\le a+b$ for positive real numbers $a$ and $b$.

The claim follows by taking $a=|x|$ and $b=|y|$.


 * $x \le 0, y \ge 0$ follows by symmetry from the previous case.