I've always understood triangle inequality as "The sum of the lengths of any two sides of a triangle is always greater or equal to the length of the remaining side", say $x, y$ and $z$ are the lengths of the sides of a triangle than $x+y ≥ z$ and in degenerate case where the vertices are collinear, $z = x+y$ and the equality holds.

But I don't understand WHY does it make sense to call $ |x+y| ≤ |x|+|y| $ the triangle inequality, where $x$ and $y$ are reals. You can't even make a triangle out of it? Can anyone tell me how to "picture" this in my head?

  • $\begingroup$ The real triangle inequality has nothing to do with triangles, you are right. If anything, in most metric spaces it has nothing to do with triangles, so "triangle inequality" is just a name. The intuition with triangles really works only with complex numbers/vectors in the plane. (but as its the same type of inequality in every metric space it makes sense to keep the name of the inequality for the general case) $\endgroup$ – Mark Aug 27 '20 at 11:19
  • $\begingroup$ You could image a “flat triangle” with vertices at $0$, $x$, $y$ on the real axis (where $x, y$ can also be negative). $\endgroup$ – Martin R Aug 27 '20 at 11:20
  • $\begingroup$ Think of it this way: if $x, y, y-x$ are elements of $\mathbb R^n$ for $n\gt 1$ they can be regarded as vectors which form a triangle. If you replace $x$ with $-x$ you get the form of triangle equality you have quoted (the absolute values don't care about signs). Now it also works with $n=1$ - so why make a new name or keep quoting a special case - the terminology is extended beyond its original meaning. This happens a lot in mathematics. $\endgroup$ – Mark Bennet Aug 27 '20 at 11:24
  • $\begingroup$ @Mark Alright, now I suppose when they do the same things with Euclidean vectors and norms, it still makes sense to call it the Triangle Inequality because you can form triangles out of the vectors and think of norm as the lengths of the sides. But I don't think you can do the same with real numbers as vectors and absolute values as the norm, can you? I do see the connection but somehow I'm fuzzy... $\endgroup$ – William Aug 27 '20 at 11:24
  • $\begingroup$ There 3 numbers and their ‘lengths’ (the absolute values) very like the lengths of the side of a triangle. The name of the inequality (including its second form) is used by metonymy. $\endgroup$ – Bernard Aug 27 '20 at 11:26

See $x$ and $y$ as vectors. For example $x = \overrightarrow{AB}$, and $y = \overrightarrow{BC}$. Then by Chasles relation, $x+y = \overrightarrow{AC}$. So the relation $$|x+y| \leq |x| + |y|$$ can be rewritten as $$|\overrightarrow{AC}| \leq |\overrightarrow{AB}| + |\overrightarrow{BC}|$$

which means that the length $AC$ of the triangle $ABC$ is less than the sum of the lenght of the two other sides.

  • $\begingroup$ I guess it's easier to picture it in $\mathbb{R}^2$ or $\mathbb{R}^3$ but picturing it in $\mathbb{R}$ is kind of difficult to me $\endgroup$ – William Aug 27 '20 at 11:35
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    $\begingroup$ In the case of $\mathbb{R}$, you just get a flat triangle, but it is the same. If you want, $|x|$ is the distance from $0$ to $x$, $|y|$ is the distance from $x$ to $x+y$, and $|x+y|$ is the distance from $0$ to $x+y$. $\endgroup$ – TheSilverDoe Aug 27 '20 at 11:40
  • $\begingroup$ Thanks, that helps a lot doe $\endgroup$ – William Aug 27 '20 at 12:01

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