Prove $|x - y|\le|x| + |y|$ (Spivak's Calculus Book) 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.
 A: Your approach is correct; do make sure that every step you have taken can also be taken in reverse. So far you have a proof that
$$\textbf{If}\quad|x-y|\leq|x|+|y|\quad\textbf{then}\quad-xy\leq|xy|,$$
but of course you want to prove the opposite implication.
A shorter proof would be to use the triangle inequality;
$$|x-y|=|x+(-y)|\leq |x|+|-y|=|x|+|y|.$$
A: 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.
A: Your solution is absolutely correct and I think that your approach is simplest and shortest in the sense that you don't need to appeal to (and prove) any lemma.
In @Servaes's approach, he appeals to this inequality $|x+y|\le |x|+|y|$, which certainly needs a proof too.
Bonus: $||x|-|y|| \le |x+y|\le |x|+|y|$.
A: You start from what you want to prove and then you square both sides, find something true and conclude.
Using this method I can prove that $$3\leq -6$$
Indeed, square both sides, you find $9\leq 36$ which is true.
What you need to add is that $a\leq b$ is indeed equivalent to $a^2\leq b^2$ whenever $a,b\geq 0$. 
That is because 


*

*if $b\geq 0$ then $$a^2\leq b^2\implies \sqrt{a^2}\leq \sqrt{b^2}=b\implies |a|\leq b\implies a\leq b$$

*if $a\geq 0$ then $a\leq b\implies a\cdot a\leq a\cdot b\leq b\cdot b$.
A: That's right to me. Also you can use the classical triangle inequality $$|x-y|=|x+(-y)|\le|x|+|-y|=|x|+|y|$$
