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Critique my proof of $\lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert, \ x,y \in \mathbb{R}$

Let $x, y \in \mathbb{R}$.

Case #1: $x = y$

$\Rightarrow \lvert x \rvert = \lvert y \rvert$. Hence, $$\lvert x + y \rvert = \lvert x \rvert + \lvert x \rvert = 2 \lvert x \rvert$$ and $$\lvert x - y \rvert = \lvert x - x\rvert = 0.$$ Thus, $\lvert x \rvert$ + $\lvert y \rvert = 2 \lvert x \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert = 2 \lvert x \rvert$.

Case #2: $x > y$

Assume for the sake of contradiction that $\lvert x \rvert + \lvert y \rvert > \lvert x + y \rvert + \lvert x - y \rvert.$ Hence, $$ \lvert x + y \rvert + \lvert x - y \rvert < \lvert x \rvert + \lvert y \rvert$$ $$\leq \lvert x \rvert + \lvert y \rvert + \lvert x - y \rvert < \lvert x \rvert + \lvert y \rvert$$ by the triangle inequality, which is a contradiction b/c $\lvert x - y \rvert > 0$. Thus, it must be that $\lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert$ in this case.

Case #3: $x < y$

WLOG, the proof is the same as case #2.

$\therefore \lvert x \rvert + \lvert y \rvert \leq \lvert x + y \rvert + \lvert x - y \rvert, \ x,y \in \mathbb{R}$.

Faith Alone
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  • Assume for the sake of contradiction. I'll edit the post so that it's more clear. – Faith Alone Nov 24 '22 at 18:49
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    Agreed with @AnneBauval, the second line in your case 2 does not follow from the first line. (If $A < B$ and $A \le C$ we cannot conclude necessarily that $C < B$.) – Torsten Schoeneberg Nov 24 '22 at 18:52
  • The sources that I'm using say that $\lvert x + y \rvert \leq \lvert x \rvert + \lvert y \rvert$. Did I make some logical leap? https://i.ytimg.com/vi/Cc7PeuDa2Zo/maxresdefault.jpg – Faith Alone Nov 24 '22 at 18:53
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    How did you get $|x|+|y|+|x-y| < |x|+|y|$ by the triangle inequality? If I'm seeing what you mean, shouldn't you have $|x-y| < |x|+|y|$ so that $|x|+|y|+|x-y| < 2|x|+2|y|$? Also, while I understand why you're trying to use those particular cases, you're going to have to bifurcate into cases such as $y>0, x<0$ eventually as well because $|y|,|x|$ etc. are involved. If you don't want to do that, you'll probably need intermediate points such as $\frac{x+y}{2}, \frac{x-y}{2}$ etc. to use with the original triangle inequality. – Sarvesh Ravichandran Iyer Nov 24 '22 at 19:10
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    @SarveshRavichandranIyer the didn't get $|x|+|y| + |x-y| < |x| +|y|$ by the triangle inequality. They got $|x+y| + |x-y| \le |x| + |y| + |x-y|$ by the T.I and $|x+y| + |x-y| < |x| + |y|$ by assumption but... as others have pointed out. Showing that $A \le C$ bt triangle equality and $A< B$ by assumption, one can not conclude $B < C$. So this prove isn't valid. – fleablood Nov 24 '22 at 19:21
  • @fleablood I see, yeah, that looks like the mistake. – Sarvesh Ravichandran Iyer Nov 24 '22 at 19:24
  • Use the triangle inequality to sho $|x+y| + |x-y| \le |x+y + x-y| = 2|x|$ and $|x+y| + |x-y| = |x+y| +|y-x| \le |x+y+y-x| = 2|y|$ and that $2\min(|x|, |y|) \le |x| + |y| \le 2\max(|x|,|y| \le |x+y| + |x-y|$. – fleablood Nov 24 '22 at 19:24
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    This is the cutest way to do it: $|x|+|y| =\frac 12(|2x| + |2y|)=\frac 12(|x+y+x-y|+|x+y+y-x|)\le \frac 12(|x+y| + |x-y| + |x+y|+|y-x|)=\frac 12(2(|x+y| + |x-y|))=|x+y| + |x-y|$. – fleablood Nov 24 '22 at 19:36

2 Answers2

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In the second case, there is a mistake because from the inequalities

$$\begin{cases}|x+y|+|x-y|<|x|+|y|\\|x+y|\leqslant|x|+|y|\end{cases}$$

you cannot deduce that

$$|x|+|y|+|x-y|<|x|+|y|\;.$$

Indeed, $\,A+B<C+D\,$ and $\,A\leqslant C+D\,$ do not imply that $\,C+D+B<C+D\,.$

For example, if $\,A=1,B=2,C=3,D=4\,,\,$ then $\,A+B<C+D\,$ and $\,A<C+D\;\;$ but $\,C+D+B\not\lt C+D\,.$

Addendum 1:

You could prove the property in the following way.

If $\,x\geqslant0\,$ and $\,y\geqslant0\,,\,$ then
$|x|+|y|=x+y=|x+y|\leqslant|x+y|+|x-y|\,.$

If $\,x<0\,$ and $\,y<0\,,\,$ then
$|x|+|y|=-x-y=-(x\!+\!y)=|x+y|\leqslant|x+y|+|x-y|\,.$

If $\,x\geqslant0\,$ and $\,y<0\,,\,$ then
$|x|+|y|=x-y=|x-y|\leqslant|x+y|+|x-y|\,.$

If $\,x<0\,$ and $\,y\geqslant0\,,\,$ then
$|x|+|y|=-x+y=-(x\!-\!y)=|x-y|\leqslant|x+y|+|x-y|\,.$

Addendum 2:

A better way to prove the property is the following one.

$\begin{align}|x|+|y|&=\left|\dfrac{(x+y)+(x-y)}2\right|+\left|\dfrac{(x+y)-(x-y)}2\right|\leqslant\\&\leqslant\dfrac{|x+y|+|x-y|}2+\dfrac{|x+y|+|x-y|}2=\\&=|x+y|+|x-y|\;.\end{align}$

Angelo
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  • Why have you downvoted my answer ? Is there any mistake ? – Angelo Nov 24 '22 at 18:55
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    I didn't read it. The question was "review my proof", not "how to prove". – Anne Bauval Nov 24 '22 at 18:57
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    No, I won't. If you want a proof review, ask for it in a separate post. As explained above, your "answer" is not an answer to this post. – Anne Bauval Nov 24 '22 at 19:00
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    you are being downvoted because you are not answering the question. You need to critique OP's prove, not hastily post your own solution just to get some internet points. – dezdichado Nov 24 '22 at 19:13
  • @Invisible, it is the first case. – Angelo Nov 24 '22 at 19:16
  • I have critiqued OP’s proof. Please, could you remove your downvotes ? – Angelo Nov 24 '22 at 20:02
  • I understand that I can't deduce $\lvert x \rvert + \lvert y \rvert + \lvert x - y \rvert < \lvert x \rvert + \lvert y \rvert$, but keep in mind that case #2 is a contradiction proof. I'm using that result to prove that there is a contradiction. – Faith Alone Nov 24 '22 at 20:05
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    @TorstenSchoeneberg, because $|x-y|\geqslant0$. – Angelo Nov 24 '22 at 20:08
  • @FaithAlone, you cannot deduce it because $,A+B<C+D,$ and $,A\leqslant C+D,$ do not imply that $,C+D+B<C+D,.$ For example, if $,A=1,B=2,C=3,D=4,,,$ then $,A+B<C+D,$ and $,A<C+D;;$ but $,C+D+B\not\lt C+D,.$ – Angelo Nov 24 '22 at 20:11
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There is a flaw in your case 2 because you used the triangle inequality the wrong way. $|x + y|+|x - y|<blah$ and $|x + y|\le|x|+|y|$ don't imply $|x|+|y|+|x - y|<blah.$

Anne Bauval
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    @FaithAlone: $\lvert 2 + (-1)\rvert + \lvert 2-(-1) \rvert < 4.5$ and yet $\lvert 2 \rvert + \lvert -1 \rvert + \lvert 2-(-1) \rvert$ is not $< 4.5$. – Torsten Schoeneberg Nov 24 '22 at 19:19
  • @Torsten Thanks! – Faith Alone Nov 24 '22 at 19:20
  • I get the concept, but in your example $\left|2\right|+\left|-1\right|+\left|2-\left(-1\right)\right| > \lvert 2 \rvert + \lvert -1 \rvert$, which still works for my contradiction proof. I've really got myself in a rabbit hole lol. – Faith Alone Nov 24 '22 at 19:41
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    Of course I cannot produce an example where the statement in your title is wrong; because that statement is true. It is just that your proof does not work, and the example shows why one conclusion in your proof, the one Anne Bauval states here, fails. – Torsten Schoeneberg Nov 24 '22 at 20:04