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Proposition: $|x+y| \leq |x| + |y|$

Proof : $(x+y)^2 = x^2 + 2xy + y^2 $

$\leq |x|^2 + 2|x||y| + |y|^2$

$= (|x| + |y|)^2$

So $(x+y)^2 \leq (|x|+|y|)^2 \Rightarrow \sqrt{(x+y)^2} \leq \sqrt{(|x| + |y|)^2} \Rightarrow |x+y| \leq |x| + |y|$

Here the fact was used that $\sqrt{(x)^2} = |x|$

I understand what is going on in this proof. No questions in that regard. I have known about this proof for awhile and I never questioned it before. I was randomly thinking about it and started overthinking the situation most likely but maybe someone can shed light on my concerns.

I feel like $x^2 + 2xy + y^2 \leq |x|^2 + 2|x||y| + |y|^2$ is just hand waved like it is well known and maybe it is. Is something like $x \leq |x|$ just assumed or does it actually need to be proven? Sorry if this question is nonsense, but I majored in math so I always think of random things like this!

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  • $\begingroup$ $u \le |u|$ for all $u \in \Bbb R$. $\endgroup$
    – user258700
    Mar 29, 2016 at 21:21
  • $\begingroup$ One of the possible definitions of $\lvert x\rvert$ is $\;\lvert x\rvert=\max(x,-x)$. If it is not the definition, it is deduced from any other definition as a simple remark. $\endgroup$
    – Bernard
    Mar 29, 2016 at 21:22
  • $\begingroup$ (For real numbers) It's hand waved but it is "well-known" $|x| \ge 0$ (with equality holding iff $x = 0$) and well known that $x = \pm|x|$ (plus if $x \ge 0$ and minus if $x \le 0$) so it's clear $x \le |x|$ (equality iff $x \ge 0$). Likewise $x^2 = |x|^2 \ge 0$ (equality iff ... well,, you know) is well known and $x > 0; y> 0\implies xy > 0; etc.$ are well known. So the result is ... elementary. You can prove it easily if you need to or take it as obvious if you want. It really depends on which context multiplication and addition axioms have been given. $\endgroup$
    – fleablood
    Mar 29, 2016 at 21:37

3 Answers 3

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It requires proof.

Let $$|x|=\begin{cases}x, & x\geq 0\\ -x, & x<0 \end{cases}$$

Suppose $x\geq 0$, then $x\leq x=|x|$. Suppose now that $x<0,$ then $|x|=-x>0>x$, thus $|x|\geq x $ for all $x$.

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The statement $x \leq |x|$ for all $x \in \mathbb{R}$ does need proving, but the proof is simple. By definition $|x|=x$ if $x \geq 0$ and $|x|=-x$ if $x<0$. If $x \geq 0$, then $x=|x|$ and thus $x \leq |x|$. On the other hand, if $x <0$, then $|x|=-x>0$, so $x<0<|x|$ and thus $x \leq |x|$. In any case we get $x \leq |x|$.

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$x^2 + 2xy + y^2 \leq |x|^2 + 2|x||y| + |y|^2$

$x^2 = |x|^2$ and $y^2 = |y|^2$

And so we can simplify this to:

$xy \leq |xy|$

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