IMO shortlist 1995/A3 solution

36th IMO 1995 shortlist

Problem A3

n > 2. x₁, x₂, ... , x_n are real numbers such that 2 ≤ x_i <= 3. Show that (x₁² + x₂² - x₃²)/(x₁ + x₂ - x₃) + (x₂² + x₃² - x₄²)/(x₂ + x₃ - x₄) + ... + (x_n-1² + x_n² - x₁²)/(x_n-1 + x_n - x₁) + (x_n² + x₁² - x₂²)/(x_n + x₁ - x₂) ≤ 2(x₁ + x₂ + ... + x_n) - 2n.

Solution

The ith term on the lhs is (x_i² + x_i+1² - x_i+2²)/(x_i + x_i+1 - x_i+2) (with cyclic subscripts). We can write this as (x_i + x_i+1 + x_i+2) - 2x_ix_i+1/(x_i + x_i+1 - x_i+2). Now (x_i - 2)(x_i+1 - 2) ≥ 0, so -2x_ix_i+1 ≤ -4(x_i + x_i+1 - 2) = -4( (x_i + x_i+1 - x_i+2) + (x_i+2 - 2) ). Hence the ith term ≤ (x_i + x_i+1 + x_i+2) - 4(1 + (x_i+2 - 2)/(x_i + x_i+1 - x_i+2) ). But the constraints on x_i imply that (x_i + x_i+1 - x_i+2) and (x_i+2 - 2) are positive and (x_i + x_i+1 - x_i+2) ≤ 4. So the ith term <= (x_i + x_i+1 + x_i+2) - 4(1 + (x_i+2 - 2)/4) = x_i + x_i+1 - 2. Hence lhs ≤ 2s - 2n.

36th IMO shortlist 1995