IMO 1999/A2 solution

IMO 1999

Problem A2

Let n >= 2 by a fixed integer. Find the smallest constant C such that for all non-negative reals x₁, ... , x_n:

∑_i<j x_i x_j (x_i² + x_j²) <= C (∑ x_i)⁴.

Determine when equality occurs.

Answer Answer: C = 1/8. Equality iff two x_i are equal and the rest zero.

Solution

By a member of the Chinese team at the IMO - does anyone know who?

(∑ x_i)⁴ = (∑ x_i² + 2 ∑_i<jx_ix_j)² ≥ 4 (∑ x_i²) (2 ∑_i<j x_ix_j) = 8 ∑_i<j ( x_ix_j ∑ x_k²) ≥ 8 ∑_i<j x_ix_j(x_i² + x_j²).

The second inequality is an equality only if n - 2 of the x_i are zero. So assume that x₃ = x₄ = ... = x_n = 0. Then for the first inequality to be an equality we require that (x₁² + x₂²) = 2 x₁x₂ and hence that x₁ = x₂. However, that is clearly also sufficient for equality.

Alternative solution by Gerhard Woeginger

Setting x₁ = x₂ = 1, x_i = 0 for i > 2 gives equality with C = 1/8, so, C cannot be smaller than 1/8.

We now use induction on n. For n = 2, the inequality with C = 1/8 is equivalent to (x₁ - x₂)⁴ ≥ 0, which is true, with equality iff x₁ = x₂. So the result is true for n = 2.

For n > 2, we may take x₁ + ... + x_n = 1, and x₁ ≤ x₂ ≤ ... ≤ x_n. Now replace x₁ and x₂ by 0 and x₁ + x₂. The sum on the rhs is unchanged and the sum on the lhs is increased by (x₁ + x₂)³ S - (x₁³ + x₂³) S - x₁x₂(x₁² + x₂²), where S = x₃ + x₄ + ... + x_n. But S is at least 1/3 (the critical case is n = 3, x_i = 1/3), so this is at least x₁x₂(x₁ + x₂ - x₁² - x₂²). This is strictly greater than 0 unless x₁ = 0 (when it equals 0), so the result follows by induction.

Comment. The first solution is elegant and shows clearly why the inequality is true. The second solution is more plodding, but uses an approach which is more general and can be applied in many other cases. At least with hindsight, the first solution is not as impossible to find as one might think. A little playing around soon uncovers the fact that one can get C = 1/8 with two x_i equal and the rest zero, and that this looks like the best possible. One just has to make the jump to replacing (x_i² + x_j²) by ∑ x_k². The solution is then fairly clear. Of course, that does not detract from the contestant's achievement, because I and almost everyone else who has looked at the problem failed to make that jump.

40th IMO 1999