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
$dfrac{1}{(a+2bk)^2}<dfrac{1}{(a+2bk)^2-lambda^2}=dfrac{1}{2lambda}left( dfrac{1}{a+2bk-lambda}-dfrac{1}{a+2bk+lambda}
ight),2lambda=2b$