Vol. 10 No 5 - Pi Mu Epsilon

PROBLEMS AND SOLUTIONS 437Solution by H. -J. Seiffert, Berlin, Germany.Since 1 + x + . . . + fln! < e* when x > 0 and n is a positive integer,we have that0 < m exp[-(p + q12 ^II. Solution by Murray S. Klamidn, University of Alberta, Ed-on,Alberta, Canada.It will be shown that the given inequality is equivalent geometrically tothe square of the volume of a tetrahedron being nonnegative.Let X, Z, and I denote, respectively, the n-dimensional vectors(x,,& ,..., x,J,(z,,z2 ,..., zy),aad(l,l, ..., 1)andlet 1 XI =x, 1 Z 1 =z, and here 1 1 1 = dn. Also let a, B, and denote the angles between Xand I, Z and I, and X and Z, respectively. The given inequality can now berewritten in the formHencem 03or equivalently*(I- cos 2 7) 2 n^$ cos2 a + nx22 cos 2 B - 2nx22 cos a cosB cos 7and finallyThe left side of the latter inequality is 36 times the square of the volume ofthe tetrahedron having X, Z, and I as three coterminal edges. There isequality if and only if the volume is 0 or equivalently if X, Z, and I arelinearly dependent. That is, if either X or Z equal 0 or k1 or there areconstants a and b such that zi = oxi + b for all i.Also solved by Paul S. Bruckman, Russell Euler, David Iny, and theProposer.874. [Fall 19951 Proposed by David Iny, Westinghouse ElectricCorporation, Baltimore, Maryland.a) Given real numbers xi and zi for 1 < i < n, prove thatb) Determine a necessary and sufficient condition for equality.11. Comment by H.-J. Seiffert, Berlin, Germany.This inequality is known. See [I], page 227, Equation 20.1.Reference1. D. S. Mitrinovit, J. E. Pdarid, and A. M. Fink, Classical and NewInequalities in Analysis, Kluwer, 1993.Also solved by Paul S. Bruckman, Joe Howard, Yoshinobu Murayoshi,and the Proposer.