Bonjour,
[tex]a+\dfrac{1}{b} =b+\dfrac{1}{c} =c+\dfrac{1}{a} \\\\a-b=\dfrac{1}{c} -\dfrac{1}{b} =\dfrac{b-c}{bc} \\\\b-c=\dfrac{1}{a} -\dfrac{1}{c} =\dfrac{c-a}{ac} \\\\c-a=\dfrac{1}{b} -\dfrac{1}{a} =\dfrac{a-b}{ab} \\\\\Longrightarrow\ (a-b)(b-c)(c-a)=\dfrac{(b-c)(c-a)(a-b)}{(abc)^2}\\ \\\Longrightarrow\ (abc)^2=1\\\\\Longrightarrow\ abc=1\ ou\ abc=-1\\[/tex]
