$ extbf{证明:}$对任意$a,bin R^+$, $dfrac{1}{sqrt{a+2b}}+dfrac{1}{sqrt{a+4b}}+dfrac{1}{sqrt{a+6b}}<dfrac{6}{sqrt{a+b}+sqrt{a+7b}}$
证明:
注意到,$dfrac{1}{sqrt{a+2kb}}<dfrac{2}{sqrt{a+(2k-1)b}+sqrt{a+(2k+1)b}}$, 故
egin{align*}
dfrac{1}{sqrt{a+2b}}+dfrac{1}{sqrt{a+4b}}+dfrac{1}{sqrt{a+6b}}
& <sumlimits_{k=1}^{3}{dfrac{2}{sqrt{a+(2k-1)b}+sqrt{a+(2k+1)b}}} \
& =sumlimits_{k=1}^{3}{dfrac{sqrt{a+ (2k+1)b}-sqrt{a+(2k-1)b}}{b}}\
&=dfrac{sqrt{a+7b}-sqrt{a+b}}{b}\
&=dfrac{6}{sqrt{a+b}+sqrt{a+7b}}
end{align*}