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  • MT【170】裂项相消

    已知$a,b>0$证明:$dfrac{1}{a+2b}+dfrac{1}{a+4b}+dfrac{1}{a+6b}<dfrac{3}{sqrt{(a+b)(a+7b)}}$

    证明:egin{align*}
    dfrac{1}{a+2b}+dfrac{1}{a+4b}+dfrac{1}{a+6b}
    & <sqrt{3}{sqrt{left(dfrac{1}{a+2b} ight)^2+left(dfrac{1}{a+4b} ight)^2+left(dfrac{1}{a+6b} ight)^2}} \
    & <sqrt{3}{sqrt{dfrac{1}{(a+b)(a+3b)}+dfrac{1}{(a+3b)(a+5b)}+dfrac{1}{(a+5b)(a+7b)}}}\
    &=sqrt{3}{sqrt{dfrac{1}{2b}left(dfrac{1}{a+b}-dfrac{1}{a+7b} ight)}}\
    &=dfrac{3}{sqrt{(a+b)(a+7b)}}.
    end{align*}

    注:这里的裂项主要是考虑到相消,一般项
    $dfrac{1}{(a+2bk)^2}<dfrac{1}{(a+2bk)^2-lambda^2}=dfrac{1}{2lambda}left( dfrac{1}{a+2bk-lambda}-dfrac{1}{a+2bk+lambda} ight),2lambda=2b$

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  • 原文地址:https://www.cnblogs.com/mathstudy/p/8961240.html
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