How do I establish the triangle inequality $|x+y| \geq |x|+|y|$ by considering all real numbers as the union of six subsets and checking the inequality on each of those subsets?

  • 3
    $\begingroup$ As the union of six subsets? Subsets of what? Also, you want $|x+y|\color{red}{\leq}|x|+|y|$. $\endgroup$
    – Pedro Tamaroff
    Apr 22 '13 at 0:48
  • $\begingroup$ We don't know, I dont think it matter what is in the six subsets, the inequality should always hold... $\endgroup$
    – kristen
    Apr 22 '13 at 0:49
  • $\begingroup$ I think you mean cases, not subsets. $\endgroup$
    – Pedro Tamaroff
    Apr 22 '13 at 0:53
  • $\begingroup$ the worksheet says subsets, but idk what the difference is... $\endgroup$
    – kristen
    Apr 22 '13 at 1:06
  • $\begingroup$ does anyone know how we would establish the general triangle inequality |x1+...+xn| less than or equal to |x1|+...+|xn| by repeating the application of the triangle inequality. $\endgroup$
    – kristen
    Apr 22 '13 at 1:08

Actually, it is the set of ordered pairs $(x,y)$ that is broken up into $6$ subsets. So we are breaking up $\mathbb{R}^2$, not $\mathbb{R}$.

The easiest case is (i) both $x$ and $y$ are $\ge 0$. The next easiest is (ii) $x\le 0$, $y\le 0$ and $x$ and $y$ not both $0$, since that's already taken care of.

Now come the mixed cases where one of $x$ or $y$ is $\lt 0$ and the other is $\gt 0$.

Each breaks up into $2$ cases. For example, with $x\gt 0$, we can have (iii) $y\lt 0$ and $|y|\ge x$ or (iv) $y\lt 0$ and $|y|\lt x$.

The case $x\lt 0$ and $y\ge 0$ also breaks up into two cases, giving a total of $6$. But we really need not worry about them, since we can take advantage of the symmetry and interchange the roles of $x$ and $y$. In view of this observation there really are only $4$ really different cases. (Actually, cases (ii) and (iv)$ are not greatly different from each other. The idea that deals with one adapts readily to deal with the other.)


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