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已知a,b,c均大于0,a+b+c=1,求证(1/a+1)(1/b+1)(1/c+1)>=64
题目详情
已知a,b,c均大于0,a+b+c=1,求证(1/a+1)(1/b+1)(1/c+1)>=64
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答案和解析
a+b+c=1≥3(abc)^1/3
abc≤1/27 1/abc≥27
(1/a+1)(1/b+1)(1/c+1)
=1/a+1/b+1/c+1/ab+1/bc+1/ac+1+1/abc≥3(1/abc)^1/3+3
(1/abc)^2/3+1/abc+1=64
所以(1/a+1)(1/b+1)(1/c+1)≥64得证
abc≤1/27 1/abc≥27
(1/a+1)(1/b+1)(1/c+1)
=1/a+1/b+1/c+1/ab+1/bc+1/ac+1+1/abc≥3(1/abc)^1/3+3
(1/abc)^2/3+1/abc+1=64
所以(1/a+1)(1/b+1)(1/c+1)≥64得证
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