若正数a,b,c满足a+b+c=1,求1/(3a+2) +1/(3b+2)+1/(3c+2)的最小值

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若正数a,b,c满足a+b+c=1,求1/(3a+2) +1/(3b+2)+1/(3c+2)的最小值

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题目内容：

若正数a,b,c满足a+b+c=1,求1/(3a+2) +1/(3b+2)+1/(3c+2)的最小值

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王老师

回答题目：2621条

解法一:
a、b、c为正实数,且a+b+c=1
故由柯西不等式得
[(3a+2)+(3b+2)+(3c+2)]*[1/(3a+2)+1/(3b+2)+1/(3c+2)]>=(1+1+1)^2
--->[3(a+b+c)+6]*[1/(3a+2)+1/(3b+2)+1/(3c+2)]>=9
--->[3×1+6]*[1/(3a+2)+1/(3b+2)+1/(3c+2)]>=9
上式两边除以9得
[1/(3a+2)+1/(3b+2)+1/(3c+2)]>=1
故取等号时,得
1/(3a+2)+1/(3b+2)+1/(3c+2)的最小值为1.
解法二:构造函数f(x)=1/(3x+2),则
f'(x)=-3(3x+2)^(-2)
f"(x)=18(3x+2)^(-3)
可见,当x>0,即x为正实数时,
f"(x)>0恒成立
故f(x)在(0,+无穷)内下凸
所以,a、b、c>0时,由琴生不等式得
f(a)+f(b)+f(c)>=3f[(a+b+c)/3]
--->1/(3a+2)+1/(3b+2)+1/(3c+2)>=3×1/[3(a+b+c)/3+2]=3×1/[3×1/3+2]=1
故1/(3a+2)+1/(3b+2)+1/(3c+2)>=1
取等号得
1/(3a+2)+1/(3b+2)+1/(3c+2)最小值为1.

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