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已知a,b,c>0,且a+2b+3c=1 求1/a+2/b+3/c 的最小值
人气:124 ℃ 时间:2020-05-05 04:36:41
解答
1/a+2/b+3/c=(a+2b+3c)/a+2(a+2b+3c)/b+3(a+2b+3c)/c=14+(2b+3c)/a+2(a+3c)/b+3(a+2b)/c=14+(2b+3c)/a+2(a+3c)/b+3(a+2b)/c=14+2b/a+3c/a+2a/b+6c/b+3a/c+6b/c=14+2b/a+2a/b+3c/a+3a/c+6b/c+6c/b>=14+4+6+12=361/a+2...
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