Prove $|x - y|\le|x| + |y|$ (Spivak's Calculus Book)

Question

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.

Answer 1

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; $$|x-y|=|x+(-y)|\leq |x|+|-y|=|x|+|y|.$$

Answer 2

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: $||x|-|y|| \le |x+y|\le |x|+|y|$.

Answer 3

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 $a\leq b\implies a\cdot a\leq a\cdot b\leq b\cdot b$.

Answer 4

That's right to me. Also you can use the classical triangle inequality $$|x-y|=|x+(-y)|\le|x|+|-y|=|x|+|y|$$

Answer 5

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.