请用数学归纳法证明一般形式的柯西不等式

无言小刀 1年前他留下的回答 已收到1个回答

半夜散粉一百人 网友

该名网友总共回答了20个问题，此问答他的回答如下：采纳率：95%

柯西不等式形式为：
(a12+a22+a32+…+an2)(b12+b22+b32+…+bn2)≥(a1b1+a2b2+a3b3+…+anbn)2
当且仅当a1/b1=a2/b2=a3/b3=…=an/bn时等号成立
设n=k时该不等式成立,则有
(a12+a22+a32+…+ak2)(b12+b22+b32+…+bk2)≥(a1b1+a2b2+a3b3+…+akbk)2
当且仅当a1/b1=a2/b2=a3/b3=…=ak/bk时等号成立
则当n=k+1时,不等式应为：
(a12+a22+a32+…+ak+12)(b12+b22+b32+…+bk+12)≥(a1b1+a2b2+a3b3+…+ak+1bk+1)2
当且仅当a1/b1=a2/b2=a3/b3=…=ak+1/bk+1时等号成立
此不等式即：
[(a12+a22+a32+…+ak2)+ak+12][(b12+b22+b32+…+bk2)+bk+12]≥[(a1b1+a2b2+a3b3+…+akbk)+ak+1bk+1]2
(a12+a22+a32+…+ak2)(b12+b22+b32+…+bk2)
+ak+12(b12+b22+b32+…+bk2)+bk+12(a12+a22+a32+…+ak2)
+ak+12bk+12≥(a1b1+a2b2+a3b3+…+akbk)2+ak+12bk+12+2ak+1bk+1(a1b1+a2b2+a3b3+…+akbk)
因为已有
(a12+a22+a32+…+ak+12)(b12+b22+b32+…+bk+12)≥(a1b1+a2b2+a3b3+…+ak+1bk+1)2
所以只须证
ak+12(b12+b22+b32+…+bk2)+bk+12(a12+a22+a32+…+ak2)+ak+12bk+12≥ak+12bk+12+2ak+1bk+1(a1b1+a2b2+a3b3+…+akbk)
即
ak+12(b12+b22+b32+…+bk2)+bk+12(a12+a22+a32+…+ak2)≥2ak+1bk+1(a1b1+a2b2+a3b3+…+akbk)
ak+12b12+ak+12b22+ak+12b32+…+ak+12bk2
+bk+12a12+bk+12a22+bk+12a32+…+bk+12ak2≥2ak+1bk+1a1b1+2ak+1bk+1a2b2+2ak+1bk+1a3b3+…+2ak+1bk+1akbk
ak+12b12+bk+12a12+ak+12b22+bk+12a22+ak+12b32+bk+12a32+…+ak+12bk2+bk+12ak2
≥2(ak+1b1)(bk+1a1)+2(ak+1b2)(bk+1a2)+2(ak+1b3)(bk+1a3)+…+2(ak+1bk)(bk+1ak)
ak+12b12+bk+12a12+ak+12b22+bk+12a22+ak+12b32+bk+12a32+…+ak+12bk2+bk+12ak2
-2(ak+1b1)(bk+1a1)-2(ak+1b2)(bk+1a2)-2(ak+1b3)(bk+1a3)-…-2(ak+1bk)(bk+1ak)≥0
[ak+12b12-2(ak+1b1)(bk+1a1)+bk+12a12]+[ak+12b22-2(ak+1b2)(bk+1a2)+bk+12a22]+…+[ak+12bk2-2(ak+1bk)(bk+1ak)+bk+12ak2]≥0
(ak+1b1-bk+1a1)2+(ak+1b2-bk+1a2)2+…+(ak+1bk-bk+1ak)2≥0
显然,若干实数的平方和一定为非复数
若等号成立,则
ak+1b1-bk+1a1=0
ak+1b2-bk+1a2=0
……
ak+1bk-bk+1ak=0
得a1/b1=a2/b2=a3/b3=…=ak+1/bk+1
所以,若柯西不等式在n=k时成立,在n=k+1时也成立
若n=1,则不等式变为
a12b12≥(a1b1)2
显然成立,所以对于n取的一切正整数,柯西不等式都成立
证明完毕,得：
柯西不等式
(a12+a22+a32+…+an2)(b12+b22+b32+…+bn2)≥(a1b1+a2b2+a3b3+…+anbn)2
当且仅当a1/b1=a2/b2=a3/b3=…=an/bn时等号成立

1年前他留下的回答

3

　　以上就是小编为大家介绍的请用数学归纳法证明一般形式的柯西不等式 的全部内容，如果大家还对相关的内容感兴趣，请持续关注上海建站网！