Let a, $b$ and $c$ be distinct nonzero real numbers such that $a$ + 1/b = $b$ + 1/c = $c$ + 1/a. Prove that |abc| = 1.

Step 1 (Formulate pairwise equations): We are given distinct nonzero real numbers $a,b,c$ such that:

We can form three separate equations:

1. $a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}⟹bc=\frac{b-c}{a-b}$

2. $b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}⟹ca=\frac{c-a}{b-c}$

3. $c-a=\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab}⟹ab=\frac{a-b}{c-a}$

Step 2 (Multiply the three equations): Multiplying the left-hand sides and right-hand sides of these three equations together:

Since $a,b,c$ are distinct, their differences are non-zero, so we can cancel them:

Step 3 (Conclusion): The value $|abc|$ is exactly 1.