IMO 1971

Let E_n = (a₁ - a₂)(a₁ - a₃) ... (a₁ - a_n) + (a₂ - a₁)(a₂ - a₃) ... (a₂ - a_n) + ... + (a_n - a₁)(a_n - a₂) ... (a_n - a_n-1). Let S_n be the proposition that E_n ≥ 0 for all real a_i.

Prove that S_n is true for n = 3 and 5, but for no other n > 2.

Solution

Take a₁ < 0, and the remaining a_i = 0. Then E_n = a₁^n-1 < 0 for n even, so the proposition is false for even n.

Suppose n ≥ 7 and odd. Take any c > a > b, and let a₁ = a, a₂ = a₃ = a₄= b, and a₅ = a₆ = ... = a_n = c. Then E_n = (a - b)³(a - c)^n-4 < 0. So the proposition is false for odd n ≥ 7.

Assume a₁ ≥ a₂ ≥ a₃. Then in E₃ the sum of the first two terms is non-negative, because (a₁ - a₃) ≥ (a₂ - a₃). The last term is also non-negative. Hence E₃ ≥ 0, and the proposition is true for n = 3.

It remains to prove S₅. Suppose a₁ ≥ a₂ ≥ a₃ ≥ a₄ ≥ a₅. Then the sum of the first two terms in E₅ is (a₁ - a₂){(a₁ - a₃)(a₁ - a₄)(a₁ - a₅) - (a₂ - a₃)(a₂ - a₄)(a₂ - a₅)} ≥ 0. The third term is non-negative (the first two factors are non-positive and the last two non-negative). The sum of the last two terms is: (a₄ - a₅){(a₁ - a₅)(a₂ - a₅)(a₃ - a₅) - (a₁ - a₄)(a₂ - a₄)(a₃ - a₄)} ≥ 0. Hence E₅ ≥ 0.

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

13th IMO 1971

© John Scholes
jscholes@kalva.demon.co.uk
8 Oct 1998
Last updated/corrected 14 Mar 03