已知,a>0,b>0,求证1/(a+b)+1/(a+2b)+……+1/(a+nb)〈n/(sqrt(a+1/2b)(a+
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已知,a>0,b>0,求证1/(a+b)+1/(a+2b)+……+1/(a+nb)〈n/(sqrt(a+1/2b)(a+(n+1)/2b)).sqrt表示开方.希望各位兄弟姐妹们帮帮忙,答得好还会再加分!最好说的稍微详细点!
∵[(a+b+……+m)/n]^2≤(aa+bb+……+mm)/n(正数均值的平方小于或等于他们平方的均值)
∴(a+b+……+m)^2≤n(aa+bb+……+mm)
运用这个重要不等式、
∴[1/(a+b)+1/(a+2b)+……+1/(a+nb)]^2<n{[1/(a+b)]^2+[1/(a+2b)]^2+……+[1/(a+nb)]^2}
<n{1/[(a+1/2b)(a+b)]+1/(a+b)(a+2b)+……+ 1/[a+(n-1)b](a+nb)}
<(n/b)*{1/(a+1/2b)-1/(a+b)+1/(a+b)-1/(a+2b)+
1/(a+3b)-1/(a+4b)+……+1/[a+(n-1)b]-1/(a+nb)}
<(n/b)*{1/(a+1/2b)-1/(a+nb)}
<(n/b)*{1/(a+1/2b)-1/(a+nb+1/2b)}
=(n/b)*{nb/[(a+1/2b)(a+nb+1/2b)]
=nn/[(a+1/2b)(a+nb+1/2b)]
<nn/{(a+1/2b)[a+(n+1)/2b)]}
即、[1/(a+b)+1/(a+2b)+……+1/(a+nb)]^2
<nn/{(a+1/2b)[a+(n+1)/2b)]}
开方即得结果,
以上有几步把分母变小、从而值变大.
∴(a+b+……+m)^2≤n(aa+bb+……+mm)
运用这个重要不等式、
∴[1/(a+b)+1/(a+2b)+……+1/(a+nb)]^2<n{[1/(a+b)]^2+[1/(a+2b)]^2+……+[1/(a+nb)]^2}
<n{1/[(a+1/2b)(a+b)]+1/(a+b)(a+2b)+……+ 1/[a+(n-1)b](a+nb)}
<(n/b)*{1/(a+1/2b)-1/(a+b)+1/(a+b)-1/(a+2b)+
1/(a+3b)-1/(a+4b)+……+1/[a+(n-1)b]-1/(a+nb)}
<(n/b)*{1/(a+1/2b)-1/(a+nb)}
<(n/b)*{1/(a+1/2b)-1/(a+nb+1/2b)}
=(n/b)*{nb/[(a+1/2b)(a+nb+1/2b)]
=nn/[(a+1/2b)(a+nb+1/2b)]
<nn/{(a+1/2b)[a+(n+1)/2b)]}
即、[1/(a+b)+1/(a+2b)+……+1/(a+nb)]^2
<nn/{(a+1/2b)[a+(n+1)/2b)]}
开方即得结果,
以上有几步把分母变小、从而值变大.
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