设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)用柯西不等式证明

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/13 14:24:15
设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)用柯西不等式证明设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/

设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)用柯西不等式证明
设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)
用柯西不等式证明

设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)用柯西不等式证明
证明:
由柯西不等式:
[(n+1)+(n+2)+...+(3n)][1/(n+1)+1/(n+2)+...+1/(3n)]>(1+1+...+1)^2=(2n)^2{注,一共有2n个1,而且等号显然不成立}
而由等差数列求和公式有:(n+1)+(n+2)+...+(3n)=n(4n+1)
于是1/(n+1)+1/(n+2)+...+1/(3n)>(4n^2)/[n(4n+1)]=4n/(4n+1)
证毕.