27th USAMO 1998

The reals x₁, x₂, ... , x_n+1 satisfy 0 < x_i < π/2 and ∑₁ⁿ⁺¹ tan(x_i - π/4) ≥ n-1. Show that ∏₁ⁿ⁺¹ tan x_i ≥ nⁿ⁺¹.

Solution

Put t_i = tan(x_i - π/4). Then (1 + t_i)/(1 - t_i) = tan(π/4 + x_i - π/4) = tan x_i. So we wish to show that ∏(1 + t_i)/(1 - t_i) ≥ nⁿ⁺¹.

The given inequality is equivalent to 1 + t_i ≥ ∑_j≠i (1 - t_j). Using the AM/GM inequality, this implies that (1 + t_i)/n >= ∏_j≠i (1 - t_j)^1/n. Hence ∏ (1 + t_i)/nⁿ⁺¹ ≥ ∏_i ∏_j≠i (1 - t_j)^1/n = ∏ (1 - t_i).

27th USAMO 1998

© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002