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+1akbkak+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=0ak+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
