2/3a+b+1/a+2b=4,求7a+4b最小值

zja2112022-10-04 11:39:541条回答

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jiuhuqixiaodui 共回答了21个问题 | 采纳率81%
依均值不等式得
7a+4b
=(1/4)·4(7a+4b)
=(1/4)·(7a+4b)[2/(3a+b)+1/(a+2b)]
=(1/4)·[2(3a+b)+(a+2b)][2/(3a+b)+1/(a+2b)]
=(1/4)[5+2(3a+b)/(a+2b)+2(a+2b)/(3a+b)]
≥5/4+(1/4)·2√[2(3a+b)/(a+2b)·2(a+2b)/(3a+b)]
=9/4.
∴2(3a+b)/(a+2b)=2(a+2b)/(3a+b)
且2/(3a+b)+1/(a+2b)=4,
即a=3/20,b=3/10时,
所求最小值为: 9/4。
1年前

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