How to prove $|x − y| ≤ |x| + |y|$, proof and reasoning

Question

Prove that, for all $x, y ∈ \mathbb{R}$, we have $|x − y| ≤ |x| + |y|$. Can I say . $|x − y|^2 ≤ (x − y)^2$, and work from there? Thank you

Answer 1

Let $x, y \in \mathbb{R}$. Consider by contradiction that $|x-y| > |x| + |y|$. Then, multiplying both sides by $|x-y|$, we have that $$|x-y|^2 > (|x| + |y|)(|x-y|)$$ $$(x-y)^2 > (|x| + |y|)(|x-y|).$$ So we have that $$ x^2 - 2xy + y^2 > (|x| + |y|)(|x-y|) > (|x| + |y|)(|x| + |y|), $$ by assumption. It follows that $$ x^2 - 2xy + y^2 > |x|^2 + 2|x||y| + |y|^2 $$ $$ x^2 - 2xy + y^2 > x^2 + 2|x||y| + y^2 $$ $$ -xy > |x||y| $$ which is a contradiction. So, we have that $|x-y| \leq |x| + |y|$, as required.