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设实数x1,x2,x3,x4,x5均不小于1,且x1·x2·x3·x4·x5=729,则max{x1x2,x2x3,x3x4,x4x5}最小=?
人气:330 ℃ 时间:2020-01-19 20:47:27
解答
x1x2+x3x4 ≥ 2√(729/x5) 即取定一个x5后,x1x2,x3x4不会都小于 √(729/x5)
x2x3+x4x5 ≥ 2√(792/x1)
√(729/x5)+√(792/x1)≥2√(729*729/x5x1)
使三个不等式等号都成立则
x1x2=x3x4=√(729/x5)
x2x3=x4x5=√(729/x1)
x1=x5
即x1=x3=x5 ,x2=x4 x1x2=x2x3=x3x4=x4x5
所以729=x1^3 *x2^2=(x1x2)^3/x2
(x1x2)^3=729*x2
x2最小为1
所以x1x2最小值为9
此时x1=x3=x5=9 x2=x4=1
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