设n为自然数,求证1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n)>4n/(4n+1)
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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
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)](/uploads/image/z/7024013-53-3.jpg?t=%E8%AE%BEn%E4%B8%BA%E8%87%AA%E7%84%B6%E6%95%B0%2C%E6%B1%82%E8%AF%811%2F%28n%2B1%29%2B1%2F%28n%2B2%29%2B1%2F%28n%2B3%29%2B...%2B1%2F%283n%29%3E4n%2F%284n%2B1%29)
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.
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.
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