$\textbf{证明:}$对任意$a,b\in R^+$, $\dfrac{1}{\sqrt{a+2b}}+\dfrac{1}{\sqrt{a+4b}}+\dfrac{1}{\sqrt{a+6b}}<\dfrac{6}{\sqrt{a+b}+\sqrt{a+7b}}$

证明:
注意到,$\dfrac{1}{\sqrt{a+2kb}}<\dfrac{2}{\sqrt{a+(2k-1)b}+\sqrt{a+(2k+1)b}}$, 故
\begin{align*}
\dfrac{1}{\sqrt{a+2b}}+\dfrac{1}{\sqrt{a+4b}}+\dfrac{1}{\sqrt{a+6b}}
& <\sum\limits_{k=1}^{3}{\dfrac{2}{\sqrt{a+(2k-1)b}+\sqrt{a+(2k+1)b}}} \\
& =\sum\limits_{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*}