Alternative triangle inequality proof

Question

I have looked everywhere for confirmation of this proof of the triangle inequality with no success.

Prove the triangle inequality:

$$\vert x + y \vert \leq \vert x \vert + \vert y \vert.$$

Proof:

Given the definition of $|x|$ to be $|x| = \max\{-x, x\}$. We know that $x \leq \vert x \vert$ and $y \leq \vert y \vert$.

So:

$$\vert x + y \vert \leq \vert \vert x \vert + \vert y \vert \vert = \vert x \vert + \vert y \vert.$$

Thus, we have proven the triangle inequality:

$$\vert x + y \vert \leq \vert x \vert + \vert y \vert.$$

What is wrong with it? If nothing is wrong with this proof, why is it not readily used as it seems to be quite concise.

Answer 1

From $x\le|x|$ and $y\le|y|$ you can deduce $$ x+y\le |x|+|y| $$ On the other hand, from $-x\le|x|$ and $-y\le|y|$ you can deduce $$ -x-y\le|x|+|y| $$ that's the same as $$ x+y\ge-(|x|+|y|) $$ The two inequalities say $$ -(|x|+|y|)\le x+y\le |x|+|y| $$ so $$ |x+y|\le|x|+|y| $$

You didn't use both conditions.

Answer 2

$$(1)\; |x|+|y|=\max (x,-x) +|y|\geq x+|y|= x+\max (y,-y)\geq x+y.$$ $$(2)\;|x|+|y|= \max (x,-x) +|y|\geq -x+|y|=-x+\max (y,-y)\geq -x-y.$$ $$ \text {Therefore }\; |x|+|y|\geq \max (x+y,-x-y)= |x+y|.$$