用数学归纳法证明一般形式的柯西不等式,

Question

用数学归纳法证明一般形式的柯西不等式,

Answer 1

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