$\begin{array}{l}
a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a}(a,b,c > 0)\\
\Rightarrow \left\{ \begin{array}{l}
a - b = \dfrac{1}{c} - \dfrac{1}{b}\\
b - c = \dfrac{1}{a} - \dfrac{1}{c}\\
c - a = \dfrac{1}{b} - \dfrac{1}{a}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a - b = \dfrac{{b - c}}{{bc}}(1)\\
b - c = \dfrac{{c - a}}{{ac}}(2)\\
c - a = \dfrac{{a - b}}{{ab}}(3)
\end{array} \right.
\end{array}$
Nhân (1),(2),(3) theo vế được :
$(a - b)(b - c)(c - a) = \dfrac{{(a - b)(b - c)(c - a)}}{{{{(abc)}^2}}}$
$\begin{array}{l}
\Rightarrow {(abc)^2} = 1 \Rightarrow abc = \pm 1\\
= > abc = 1 (a,b,c>0 => abc>0)
\end{array}$
Do đó: $\sqrt 7 abc = \sqrt 7 $