15th Balkan 1998

x_i are distinct positive reals satisfying x₁ < x₂ < ... < x_2n+1. Show that x₁ - x₂ + x₃ - x₄ + ... - x_2n + x_2n+1 < (x₁ⁿ - x₂ⁿ + ... - x_2nⁿ + x_2n+1ⁿ)^1/n.

Solution

It is easier to prove the more general result where the powers and root use any m > 1, instead of n: x₁ - x₂ + x₃ - x₄ + ... - x_2n + x_2n+1 < (x₁^m - x₂^m + ... - x_2n^m + x_2n+1^m)^1/m. The reason is that it is then essentially sufficient to prove it for n = 1. For having established it for n = 1 we have an almost trivial induction (it is much easier than it looks). Suppose it is true for n, then x₁ - x₂ + x₃ - x₄ + ... - x_2n + x_2n+3 < (x₁^m - x₂^m + ... - x_2n^m + x_2n+1^m)^1/m - x_2n+2 + x_2n+3. Now put y = (x₁^m - x₂^m + ... - x_2n^m + x_2n+1^m)^1/m. Since (x₁^m - x₂^m + ... - x_2n^m + x_2n+1^m) = x₁^m + (x₃^m - x₂^m + (x₅^m - x₄^m) + ... + (x_2n+1^m - x_2n^m) and each term is positive, y is positive. But x_2n+2^m - y^m = (x_2n+2^m - x_2n+1^m) + (x_2n^m - x_2n-1^m) + ... + (x₂^m - x₁^m) and again each term is positive so y < x_2n+2. So we can now apply the theorem for n = 1 to get y - x_2n+2 + x_2n+3 < (y^m - x_2n+2^m + x_2n+3^m)^1/m, which gives the result for n+1.

So it remains to prove the case n = 1. In other words we must show that for 0 < x < y < z we have x - y + z < (x^m - y^m + z^m)^1/m. Put f(x) = x - y + z - (x^m - y^m + z^m)^1/m. Then f '(x) = 1 - (x^m/(x^m - y^m + z^m))^1-1/m. But x^m < x^m - y^m + z^m, so f '(x) > 0 for 0 < x < y. But f(y) = 0, so f(x) < 0 for 0 < x < y, which is the required result.

15th Balkan 1998

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