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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

omm
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  • How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved? –  Dec 18 '18 at 04:24
  • In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$. – Matt Samuel Dec 18 '18 at 04:26
  • so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|... – omm Dec 18 '18 at 04:28
  • @omm: Yes. You are right. –  Dec 18 '18 at 07:04

1 Answers1

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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.

alex
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