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:\begin{align*}
\dfrac{1}{a+2b}+\dfrac{1}{a+4b}+\dfrac{1}{a+6b}
& <\sqrt{3}{\sqrt{\left(\dfrac{1}{a+2b}\right)^2+\left(\dfrac{1}{a+4b}\right)^2+\left(\dfrac{1}{a+6b}\right)^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}\right)}}\\
&=\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
$\dfrac{1}{(a+2bk)^2}<\dfrac{1}{(a+2bk)^2-\lambda^2}=\dfrac{1}{2\lambda}\left( \dfrac{1}{a+2bk-\lambda}-\dfrac{1}{a+2bk+\lambda}\right),2\lambda=2b$