定理1 如果a_1,a_2,…,a_n均为正数,那么 multiply from i=1 to n(1+a_i)≥(1+multiply from i=1 to n(a_i)~(1/2))~n。证明∵a_1,a_2,…,a_n均为正数,依据算术平均值与几何平均值的不等关系,有sum from i=1 to n a_i≥n multiply from i=1 to n a_i~(1/n)=C_n~1 multiply from i=1 to n a_i~(1/n) 从a_1,a_2,…,a_n n个元素中,每次选取两个不同元素的积的个数是1/2n(n-)=C_n~2。这C_n~2个积的连乘积里含a_1,a_2,…,a_n各元素的幂指数显然都是C_(n-1)~1=n-1。仍然依据算术平均值与几何 Theorem 1 If a_1, a_2, ..., a_n are all positive, then multiply from i=1 to n(1+a_i)≥(1+multiply from i=1 to n(a_i)~(1/2))~ n. Prove that ∵a_1, a_2, ..., a_n are all positive numbers. According to the inequality relation between the arithmetic mean value and the geometric mean value, there is sum from i=1 to n a_i≥n multiply from i=1 to n a_i~(1/ n)=C_n~1 multiply from i=1 to n a_i~(1/n) From a_1,a_2,...,a_n n elements, the number of products for selecting two different elements at a time is 1/2n ( n-)=C_n~2. The product of the powers of a_1, a_2, ..., a_n in the continuous product of C_n~2 products is obviously C_(n-1)~1=n-1. Still based on arithmetic mean and geometry