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若正数abc满足a+b+c=1求1/3a+2+1/3b+2+1/3c+2的最小值
题目详情
若正数abc满足a+b+c=1
求1/3a+2 +1/3b+2 +1/3c+2的最小值
求1/3a+2 +1/3b+2 +1/3c+2的最小值
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答案和解析
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.
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.
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