Mixmatch

Equating pairs of the given expressions in turn, we have that

• (i) $a + \dfrac{1}{b} = b + \dfrac{1}{c} \Longrightarrow a - b = \dfrac{1}{c} - \dfrac{1}{b} \Longrightarrow a - b = \dfrac{b - c}{bc}$ ,

.

• (ii) $a + \dfrac{1}{b} = c + \dfrac{1}{a} \Longrightarrow a - c = \dfrac{1}{a} - \dfrac{1}{b} \Longrightarrow a - c = \dfrac{b - a}{ab},$

.

• (iii) $b + \dfrac{1}{c} = c + \dfrac{1}{a} \Longrightarrow b - c = \dfrac{1}{a} - \dfrac{1}{c} \Longrightarrow b - c = \dfrac{c - a}{ac}.$

Multiplying the resulting equations (i), (ii) and (iii) yields that

$(a - b)(a - c)(b - c) = \dfrac{(b - c)(b - a)(c - a)}{(abc)^{2}}.$

Now with $a,b,c$ distinct reals, (and clearly $a,b,c \ne 0$ ), we can cancel terms to find that

$(abc)^{2} = 1 \Longrightarrow \boxed{|abc| = 1}.$

(One solution example is $(a,b,c) = (1, -\frac{1}{2}, -2)$ .)

Take 1 as an example.