摘 要: | ![]() 新教材第二册上P30复习参考题第 8题 :已知a >b >c ,求证 :1a -b+ 1b -c+ 1c -a>0 .现对该题进行如下推广 .推广 1 若a >b >c ,m ,n均为正数 ,则 ma -b+ nb -c+ (m +n) 2c-a ≥ 0 .证 ∵ (a -c ) ( ma -b + nb -c) =m(a -b +b -c)a -b + n(a -b +b -c)b -c =m +n + [m·b -ca -b+n·a -bb -c]≥m +n + 2mn =(m +n) 2 ,故 :ma -b+ nb -c+ (m +n) 2c -a ≥ 0 .推广 2 若a1>a2 >a3 >… >an >an + 1,则1a1-a2+ 1a2 -a3+… + 1an-an + 1+ n2an + 1-a1≥ 0证 利用柯西不等式 .∵ (an -an + 1) ( 1a1-a2+ 1a2 -a3+… +1an-an + 1) =[(a1-a2 ) …
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