已知a,b,c为R,满足a+2b+3c=1,则1/a +1/2b +1/3c最小值为?

问题描述:

已知a,b,c为R,满足a+2b+3c=1,则1/a +1/2b +1/3c最小值为?

1/a +1/2b +1/3c
=(1/a +1/2b +1/3c)(a+2b+3c)
=1+a/2b+a/3c+2b/a+1+2b/3c+3c/a+3c/2b+1
=3+(a/2b+2b/a)+(a/3c+3c/a)+(3c/2b+2b/3c)
≥3+2√(a/2b×2b/a)+√(a/3c×3c/a)+2√(3c/2b+2b/3c)
-3+2+2+2
=9
答案:最小值为9