若不等式1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)>a/24 对一切正整数 都成立,求正整数a的最大值,并证明
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![若不等式1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)>a/24 对一切正整数 都成立,求正整数a的最大值,并证明](/uploads/image/z/657847-55-7.jpg?t=%E8%8B%A5%E4%B8%8D%E7%AD%89%E5%BC%8F1%2F%28n%2B1%29%2B1%2F%28n%2B2%29%2B1%2F%28n%2B3%29%2B.%2B1%2F%283n%2B1%29%3Ea%2F24+%E5%AF%B9%E4%B8%80%E5%88%87%E6%AD%A3%E6%95%B4%E6%95%B0+%E9%83%BD%E6%88%90%E7%AB%8B%2C%E6%B1%82%E6%AD%A3%E6%95%B4%E6%95%B0a%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%2C%E5%B9%B6%E8%AF%81%E6%98%8E)
若不等式1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)>a/24 对一切正整数 都成立,求正整数a的最大值,并证明
若不等式1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)>a/24 对一切正整数 都成立,求正整数a的最大值,并证明
若不等式1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)>a/24 对一切正整数 都成立,求正整数a的最大值,并证明
f(n)=1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)
f(n+1)-f(n)=1/(3n+2)+1/(3n+3)+1/(3n+4)-1/(n+1)
=2/(3n+2)(3n+3)(3n+4)>0
f(n)递增
所以f(n)最小值为f(1)=13/12
1/(n+1)+1/(n+2)+1/(n+3)+.+1/(3n+1)>a/24 对一切正整数 都成立
所以a/24
a<26
所以a的最大正整数为25
柯西不等式:(a1^2+a2^2+...+an^2)(b1^2+b2^2+..bn^2)>=(a1b1+a2b2+..anbn)^2
当a1/b1=a2/b2=...an/bn时,取等号.
所以:
[1/(n+1)+1/(n+2)+1/(n+3)+......+1/(3n+1)](n+1+n+2+n+3+....+3n+1)>(1+1+..+1)^2=(2n)^2=4n^2...
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柯西不等式:(a1^2+a2^2+...+an^2)(b1^2+b2^2+..bn^2)>=(a1b1+a2b2+..anbn)^2
当a1/b1=a2/b2=...an/bn时,取等号.
所以:
[1/(n+1)+1/(n+2)+1/(n+3)+......+1/(3n+1)](n+1+n+2+n+3+....+3n+1)>(1+1+..+1)^2=(2n)^2=4n^2
(n+1+n+2+...3n+1)=(3n+1+n+1)*2n/2=(4n+2)n (一共2n项)
[1/(n+1)+1/(n+2)+1/(n+3)+......+1/(3n+1)]>4n^2/[(4n+2)n]=2n/(2n+1)>=a/24
2n/(2n+1)=[2n+1-1]/(2n+1)=1-1/(2n+1) n=1时,1/(2n+1)最大,-1/(2n+1)最小,1-1/(2n+1)最小,
2n/(2n+1)最小值为:2/(2+1)=2/3
要使用2n/(2n+1)>=a/24恒成立,必须a/24<=2/3
即: a<=16 a的最大值为16
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