一个幂指不等式猜想的部分肯定解决  

Partly Affirmative Answer to a Power Exponent Inequality Conjecture

作　　者：刘燕红[1] 成刚[1] 周淑情[1] 许玲娜[1] 徐耀[1] 倪仁兴[1] 

机构地区：[1]绍兴文理学院数学系,浙江绍兴312000

出　　处：《绍兴文理学院学报》2008年第8期27-29,共3页Journal of Shaoxing University

基　　金：绍兴文理学院校级教改立项资助项目(070204);浙江省教育厅科研计划重点资助项目(20061154);浙江省自然科学基金资助项目(Y606717)

摘　　要：倪仁兴和张森国于2002年提出了下面一个幂指不等式猜想:对满足0<a1 a2…an的{ak},不等式sum from k=1 to n akan-k+1≤sum from j=1 to n akjamj≤sum from k=1 to n akak成立,其中{k1,k2,…,kn}与{m1,m2,…,mn}是{1,2,…,n}的任意两个排列.此猜想已被收入匡继昌教授于2004年出版的专著——《常用不等式》(第三版),并把它列为至今不等式问题中未解决的152个猜想的第25个,可见,解决此猜想是有意义的和十分重要的.文章的目的是证明此猜想的右边不等式是成立的,从而部分肯定地解决此猜想.In 2002, Ni Renxing and Zhang Senguo put forward the following conjecture that power exponent inequality n∑k=1akaa-k＋1≤a∑j=1akj^amj≤a∑k=1ak^ak is true when {ak}satisfies0〈a1≤a2≤…an where{k1,k2,…,kn}and{m1,m2,…,mn} are the two arbitrary arrangements of { 1,2,…… n } .The conjecture had been collected in professor Kuang Jichang＇s Common Use Inequalities Commonly Used published in 2004, and it was also be set the 25th one between 152 conjectures on inequality that have not been solved. So the conjecture is very significant and important. The right side inequality of this conjecture had been proved to be true , thereby we can partly affirmative answer the conjecture.

关 键 词：猜想 排列 幂指不等式 部分肯定回答 

分 类 号：O178[理学—数学]