我们知道,柯西不等式:a_i,b_i∈R,则sum from i=1 to n a_i~2 sum from i=1 to n b_i~2≥(sum from i=1 to n a_ib_i)~2……(1)当且仅当a_i=kb_i(i=1,2,…,n)不等式等号成立。它可以作如下变形: 由(1)得(sum from i=1 to n a_i~2 sum from i=1 to n b_i~2)~(1/2)≥sum from i=1 to n a_ib_i,添项变为sum from i=1 to n a_i~2+2 (sum from i=1 to n a_i~2 sum from i=1 to n b_i~2)~(1/2)+sum from i=1 to n b_i~2≥sum from i=1 to n a_i~2+2sum from i=1 to n a_ib_i+sum from i=1 to n b_i~2,或sum from i=1 to n a_i~2-2 (sum from i=1 to n a_i~2 sum from i=1 to n b_i~2)~(1/2)+sum from i=1 to n b_i~2≤sum from i=1 to n a_i~2-2 sum from i=1 to n a_i b_i+sum from i=1 to n b_i~2,分别配方,并开方转 We know that Cauchy inequality: a_i, b_i∈R, then sum from i=1 to n a_i~2 sum from i=1 to n b_i~2≥(sum from i=1 to n a_ib_i)~2......( 1) If and only if the a_i=kb_i(i=1,2,...,n) inequality sign holds. It can be modified as follows: From (1) (sum from i=1 to n a_i~2 sum from i=1 to n b_i~2)~(1/2)≥sum from i=1 to n a_ib_i, add The item becomes sum from i=1 to n a_i~2+2 (sum from i=1 to n a_i~2 sum from i=1 to n b_i~2)~(1/2)+sum from i=1 to n b_i~2≥sum from i=1 to n a_i~2+2sum from i=1 to n a_ib_i+sum from i=1 to n b_i~2, or sum from i=1 to n a_i~2-2 ( Sum from i=1 to n a_i~2 sum from i=1 to n b_i~2)~(1/2)+sum from i=1 to n b_i~2≤sum from i=1 to n a_i~2- 2 sum from i=1 to n a_i b_i+sum from i=1 to n b_i~2, respectively formula, and turn square turn