设a,b,c大于零,且a+2b+3c=3,则1⼀a+1⼀2b+1⼀3c的最小值为

2024-11-23 08:54:36
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回答(1):

(1/a+1/2b+1/3c)(a+2b+3c)

=3+(2b/a+a/2b)+(3c/a+a/3c)+(3c/2b+2b/3c)

≥3+2+2+2

=9

∵  a+2b+3c=3

∴  1/a+1/2b+1/3c≥3

即:1/a+1/2b+1/3c的最小值为3