§ Barycentric subdivision: edge length decreases

For a edge EE, subdiving the edges into two at the center produces two edges both 1/21/2 the original length. Given a triangle TT, we wish to prove that subdividing the triangle by joining the barycenter to the vertices reduces edge length by 2/32/3 of the maximum length. More generally, we wish to show that the edge length decreases to n/(n+1)n/(n+1) of the largest length for an nn dimensional figure. The barycenter is at the location b1/nivib \equiv 1/n \sum_i v_i. The distance from a vertex vkv_k is vjb=vk(1/nivi)=1/n(ivkvi)||v_j - b|| = ||v_k - (1/n \sum_i v_i)|| = ||1/n(\sum_i v_k - v_i)||. By Cauchy Schwarz, we have that vjb1/nivkvi||v_j - b|| \leq 1/n ||\sum_i v_k - v_i||. One of the terms, where k=ik = i will be zero, and the other (n-1) terms are at most ll, the length of the longest edge. This gives vjb(n1)l/n||v_j - b|| \leq (n-1)l/n, hence the edge length decreases by a factor of (n1)/n(n-1)/n.