triangle inequality

Let $(X,d)$ be a metric space. The triangle inequality statesthat for any three points $x,y,z\in X$ we have

$$d(x,y)\le d(x,z)+d(z,y).$$

The name comes from the special case of ${ℝ}^n$ with the standardtopology, and geometrically meaning that in any triangle, the sum ofthe lengths of two sides is greater (or equal) than the third.

Actually, the triangle inequality is one of the properties that definea metric, so it holds in any metric space. Two important cases are$ℝ$ with $d(x,y)=|x-y|$ and $ℂ$ with $d(x,y)=\parallel x-y\parallel$ (herewe are using complex modulus, not absolute value).

There is a second triangle inequality, sometimes called thereverse triangle inequality, which also holds in any metricspace and is derived from the definition of metric:

$$d(x,y)\ge |d(x,z)-d(z,y)|.$$

In Euclidean geometry, this inequality is expressed by saying that eachside of a triangle is greater than the difference of the other two.

The reverse triangle inequality can be proved from the first triangleinequality, as we now show.

Let $x,y,z\in X$ be given. For any $a,b,c\in X$, from the firsttriangle inequality we have:

$$d(a,b)\le d(a,c)+d(c,b)$$

and thus (using $d(b,c)=d(c,b)$ for any $b,c\in X$):

$$d(a,c)\ge d(a,b)-d(b,c)$$(1)

and writing (1) with $a=x,b=z,c=y$:

$$d(x,y)\ge d(x,z)-d(z,y)$$(2)

while writing (1) with $a=y,b=z,c=x$ we get:

$$d(y,x)\ge d(y,z)-d(z,x)$$

or

$$d(x,y)\ge d(z,y)-d(x,z);$$(3)

from (2) and (3), using the properties of theabsolute value, it follows finally:

$$d(x,y)\ge \left|d(x,z)-d(z,y)\right|$$

which is the second triangle inequality.