Question

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

Answer 1

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。

Answer 2

解：
∵a+b+c=1.
∴9=(3a+2)+(3b+2)+(3c+2).
∴由题设及柯西不等式可得：
9×原式
=[(3a+2)+(3b+2)+(3c+2)]×[1/(3a+2)+1/(3b+2)+1/(3c+2)]≥
(1+1+1)²=9
即原式≥1.等号仅当a=b=c=1/3时取得。
∴原式min=1.

Answer 3

由柯西不等式
1/(3a+2)+1/(3b+2)+1/(3c+2)
≥[(1+1+1)^2]/(3a+3b+3c+6)
=1
当且仅当a=b=c=1/3时，等号成立
故原式的最小值为1