Let n >= 2 by a fixed integer. Find the smallest constant C such that for all non-negative reals x1, ... , xn:
∑i<j xi xj (xi2 + xj2) <= C (∑ xi)4.
Determine when equality occurs.
Answer Answer: C = 1/8. Equality iff two xi are equal and the rest zero.
Solution
By a member of the Chinese team at the IMO - does anyone know who?
(∑ xi)4 = (∑ xi2 + 2 ∑i<jxixj)2 ≥ 4 (∑ xi2) (2 ∑i<j xixj) = 8 ∑i<j ( xixj ∑ xk2) ≥ 8 ∑i<j xixj(xi2 + xj2).
The second inequality is an equality only if n - 2 of the xi are zero. So assume that x3 = x4 = ... = xn = 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, xi = 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 + ... + xn = 1, and x1 ≤ x2 ≤ ... ≤ xn. 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 + ... + xn. But S is at least 1/3 (the critical case is n = 3, xi = 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 xi equal and the rest zero, and that this looks like the best possible. One just has to make the jump to replacing (xi2 + xj2) by ∑ xk2. 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.
(C) John Scholes
jscholes@kalva.demon.co.uk
3 Sep 1999
Last corrected/updated 19 Aug 03