IMO 1981 SL 3

Origin: CAN

Problem

Find the minimum value of max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g) subject to the constraints (i) a, b, c, d, e, f, g \geq0, (ii) a + b + c + d + e + f + g = 1.

Solution

Denote max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g) by p. We have (a + b + c) + (c + d + e) + (e + f + g) = 1 + c + e \leq3p, which implies that p \geq1/3. However, p = 1/3 is achieved by taking (a, b, c, d, e, f, g) = (1/3, 0, 0, 1/3, 0, 0, 1/3). Therefore the answer is 1/3. Remark. In fact, one can prove a more general statement in the same way. Given positive integers n, k, n \geqk, if a1, a2, . . . , an are nonnegative real numbers with sum 1, then the minimum value of maxi=1,...,n−k+1{ai + ai+1 + \cdot \cdot \cdot + ai+k−1} is 1/r, where r is the integer with k(r −1) < n \leqkr.