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链接：bbs.pep.com.cn/forum.php?mod=viewthread&tid=2875409
题目：已知a.b大于0,求证1/(a+2b)+1/(a+4b)+1/(a+6b)<3/根号(a+b)(a+7b)
\begin{align*}
\left(\frac1{a+2b}+\frac1{a+4b}+\frac1{a+6b}\right)^2&<3\left(\frac1{(a+2b)^2}+\frac1{(a+4b)^2}+\frac1{(a+6b)^2}\right)\\
&<3\left(\frac1{(a+b)(a+3b)}+\frac1{(a+3b)(a+5b)}+\frac1{(a+5b)(a+7b)}\right)\\
&=\frac3{2b}\left(\frac1{a+b}-\frac1{a+3b}+\frac1{a+3b}-\frac1{a+5b}+\frac1{a+5b}-\frac1{a+7b}\right)\\
&=\frac9{(a+b)(a+7b)}.
\end{align*} |
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