因为：a-d=(a-b)+(b-c)+(c-d)≥3*[(a-b)(b-c)(c-d)]^(1/3)＞0 又， 1/(a-b)+1/(b-c)+1/(c-d)≥3*[1/(a-b)(b-c)(c-d)]^(1/3) 上述两式取等号的条件均为：a-b=b-c=c-d 所以： [1/(a-b)+1/(b-c)+1/(c-d)]*(a-d)≥3*[(a-b)(b-c)(c-d)]^(1/3)*3*[1/(a-b)(b-c)(c-d)]^(1/3) =9 即，[1/(a-b)+1/(b-c)+1/(c-d)]*(a-d)的最小值为9
答：(1) n=33时,2^3=2(3+1),不等式成立. (2) 假设n=k时,2^k≥2(k+1),则当n=k+1时, 2^(k+1)=2·2^k≥2·2(k+...详情>>
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