Russian 00/6 solution

Russian 2000

Problem 6

We have -1 < x₁ < x₂ < ... < x_n < 1 and y₁ < y₂ < ... < y_n such that x₁ + x₂ + ... + x_n = x₁¹³ + x₂¹³ + ... + x_n¹³. Show that x₁¹³y₁ + x₂¹³y₂ + ... + x_n¹³y_n < x₁y₁ + ... + x_ny_n.

Solution

Note that if x_i > 0, then x_i¹³ < x_i and if x_i < 0, then x_i¹³ > x_i. So not all the x_i can be non-negative and not all can be non-positive. Hence x₁ < 0 < x_n.

Put z_i = x_i - x_i¹³. Then z₁ < z₂ < z₃ < ... < z_n. So if z_i + z_i+1 + ... + z_n ≤ 0 for i > 1, then z_i must be negative and hence all of z₁, z₂, ... , z_i-1 must be negative and hence z₁ + z₂ + ... + z_n < 0. Contradiction. hence z_i + z_i+1 + ... + z_n > 0 for i > 1.

Now we have ∑ x_iy_i - ∑ x_iy_i¹³ = ∑ y_iz_i = y₁(z₁ + z₂ + ... + z_n) + (y₂ - y₁)(z₂ + ... + z_n) + (y₃ - y₂)(z₃ + ... + z_n) + ... + (y_n - y_n-1)z_n > 0.

Russian 2000