TITLE: Prove $|x - y|\le|x| + |y|$ (Spivak's Calculus Book)
QUESTION [7 upvotes]: I was doing this problem and I got stuck so much time, and then I conclude this.
$$|x - y|\le|x| + |y| .$$
Then, square both sides
$$|(x - y)(x - y)|\le(|x| + |y|)².$$
Since $(x - y)²$ is always positive, as well as $x²$ and $y²$, we get
$$(x - y)²\le x² + 2|xy| + y².$$
Develop the product
$$x² -2xy + y² \le x² + 2|xy| + y²$$
$$-2xy\le2|xy|$$
$$-xy\le|xy|$$
And this is true.
I'd like to know if I did this correctly and If there's a smaller/easier proof because the problem is doing by a very short proof.

REPLY [2 votes]: Your proof is mostly correct, but you should mention that, for $a\ge0$ and $b\ge0$,

$a\le b$ if and only if $a^2\le b^2$

and that you can apply this because $|x-y|\ge0$ and $|x|+|y|\ge0$. Also you should say that the inequalities you write are equivalent to each other.
In the following steps, each line is equivalent to the one below it; also we use $|a|^2=a^2$ and that adding equal terms to both sides of an inequality doesn't change it.
\begin{align}
& |x-y|\le|x|+|y| \\[4px]
& |x-y|^2\le(|x|+|y|)^2 \\[4px]
& x^2-2xy+y^2\le x^2+2|x|\,|y|+y^2 \\[4px]
& {-}2xy\le |2xy|
\end{align}
Since the last inequality is true (because $-a\le|a|$ for every $a$), we end the proof.