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最值问题中,有一类在给定条件下求最大值的问题,可用构造条件的方法求解。现介绍如下: 有关定理(柯西不等式): 对于任意实数a_i,b_i(i=1,2,…n),有:(a1b1+a2b2+…+a_nb_n)~2≤(a~21+a~22+…+a~2n)·(b~21+b~22+…+b~2n).其中,当且仅当a_i=kbi时取等号。 由柯西不等式,易得如下推论: 如果:(a~21+a~22+…+a~2n=S2(常数S>0) b~21+b~22+…+b~2n=t~2(常数t>0) 那么:a1b1+a2b2+…+a_nb_n≤S·t,当且仅当a_i/b_i=s/t(i=1,2,…,n)时,取等号,即a1b1+a2b2+…+a_nb_n有最大值s·t. 例1:已知:a2+b2+c2=1,求的最大值。 分析:为了利用推论,必须
In the problem of the maximum value, there is a problem of finding the maximum value under a given condition, which can be solved by the method of constructing conditions. The following is introduced: Theorems (Caucy inequality): For any real number a_i,b_i (i=1,2,...n), there are: (a1b1+a2b2+...+a_nb_n)~2≤(a~21+a~22) +...+a~2n)·(b~21+b~22+...+b~2n). Among them, the equal sign is used if and only if a_i=kbi. From Cauchy inequality, it is easy to get the following inference: If: (a~21+a~22+...+a~2n=S2 (constant S>0) b~21+b~22+...+b~2n=t~ 2 (constant t>0) Then: a1b1+a2b2+...+a_nb_n≤S·t, if and only if a_i/b_i=s/t(i=1,2,...,n), take the equal sign, that is, a1b1 +a2b2+...+a_nb_n has a maximum value of s·t. Example 1: Known: a2+b2+c2=1, the maximum value of the search. Analysis: In order to use inferences, you must