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.
