\(a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}\)
\(\Rightarrow a-b=\dfrac{1}{c}-\dfrac{1}{b}=\dfrac{b-c}{bc}\)(1)
\(\Rightarrow b-c=\dfrac{1}{a}-\dfrac{1}{c}=\dfrac{c-a}{ac}\)(2)
\(\Rightarrow c-a=\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{b-a}{ab}\)(3)
Nhân vế theo vế của (1);(2);(3) ta được :
\(\left(a-b\right)\left(b-c\right)\left(c-a\right)=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(abc\right)^2}\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)-\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(abc\right)^2}=0\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)\left[1-\dfrac{1}{a^2b^2c^2}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b=c\\a^2b^2c^2=1\end{matrix}\right.\)(đpcm)
