Competition problem — Algebra — Olympiad Maths Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Distinct real numbers $a, b, c$ satisfy the condition $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$ . Find all possible values of the product $abc$ :
1) for all real $a, b, c$ ;
2) for positive real $a, b, c$ ?

Solution

a) See problem 9.5

b) We will show that there are no such positive numbers that satisfy the condition of the problem. Without loss of generality we suppose that $a > b$ , then from the equality $(a - b) = \frac{b - a}{b - c}$ we conclude that $b > c$ . From the condition $b - c = \frac{c - a}{a - c}$ , we can get a contradiction: $c > a > b > c$ .

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