解法一:
a、b、c為正實(shí)數(shù),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
上式兩邊除以9得
[1/(3a+2)+1/(3b+2)+1/(3c+2)]>=1
故取等號(hào)時(shí),得
1/(3a+2)+1/(3b+2)+1/(3c+2)的最小值為1.
解法二:構(gòu)造函數(shù)f(x)=1/(3x+2),則
f'(x)=-3(3x+2)^(-2)
f"(x)=18(3x+2)^(-3)
可見(jiàn),當(dāng)x>0,即x為正實(shí)數(shù)時(shí),
f"(x)>0恒成立
故f(x)在(0,+無(wú)窮)內(nèi)下凸
所以,a、b、c>0時(shí),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
取等號(hào)得
1/(3a+2)+1/(3b+2)+1/(3c+2)最小值為1.