The Triangle Inequality and Its Applications in the Relative Metric Space *

Let C be a plane convex body. For arbitrary points , denote by , n a b E  ab the Euclidean length of the line-segment . Let be a longest chord of C parallel to the line-segment . The relative distance between the points and is the ratio of the Euclidean distance between and b to the half of the Euclidean distance between and . In this note we prove the triangle inequality in with the relative metric , and apply this inequality to show that ab 1 1 a b 1 a ab a E  , C d a b

Proof.By the properties of affine transformation, we may assume that the triangle xyz formed by the points , , x y z is a regular triangle.Let 1 1 2 2 , x y x z , and 1 2 be the affine diameters of C parallel to xy, xz, yz re-z y Copyright © 2013 SciRes.OJDM Z. J. SU ET AL. 128 spectively, and let 1 1 Since xyz is a regular triangle, by the definition of relative distance, we need to prove the following inequality.
Take the lines 1 x u and 2 x v through the points 1 x and 2 x , respectively, such that they are parallel to 1 2 z y , where (resp.) is the intersection point of the lines u v ) and 1 2 y z .Denote by  the relative distance between the points 1 x and .(See Figure 1) Since 1 2 is an affine diameter of C , we obtain The following equality is obvious.
Adding all these triangle inequalities, we obtain that , , , , are the midpoints of the sides , respectively.By the properties of affine transformation we suppose without loss of generality that is a square.Let 1 2 be the vertices of P between 1 and .Let i v , be the perpendicular projection of i onto the line segment i n   , be the perpendicular projection of i onto the line segment .(See Figure 3) According to Lemma 1, we obtain that , , , , , Adding all these inequalities, we have , , a b E  ab the Euclidean length of the line-segment .Let be a longest chord of C parallel to the line-segment .The relative distance between the points and is the ratio of the Euclidean distance between and b to the half of the Euclidean distance between and .In this note we prove the triangle inequality in with the perimeter of the convex polygon measured in the metric .In addition, we prove that every convex hexagon has two pairs of consecutive vertices with relative distances at least 1.Relative Distance; Triangle Inequality; Hexagon We use some definitions from [1].For arbitrary points , denote by ab the line-segment connecting the points a and b, by , n a b E  ab the Euclidean length of the line-segment ab, and by ab the straight line passing through the points a and .Let 1 1 be a longest chord of C parallel to ab.The C-distance C between the points is defined by the ratio of Figure 1.The figure of Lemma 1.
There are two different configurations in this case, as shown in Figure 6.In (1) of Figure 6, since a and d are midpoints of the sides xw and of P , respectively, Case 1.The parallelogram P has two sides, each of which contains exactly two vertices of H .This case contains two different configurations, as shown in Figure 4. We first consider (1) in Figure 4.
. We denote by n * Su's research was partially supported by National Natural Science Foundation of China (11071055) and NSF of Hebei Province (A2013-205089).Shen's research was partially supported by NSF (CNS 0835834, DMS 1005206) and Texas Higher Education Coordinating Board (ARP 003615-0039-2007).
Denote by H the given convex hexagon.