IMO 1977 LL CZS7

Origin: CZS

Problem

Prove the following assertion: If c1, c2, . . . , cn (n \geq2) are real numbers such that (n −1)(c2 1 + c2 2 + \cdot \cdot \cdot + c2 n) = (c1 + c2 + \cdot \cdot \cdot + cn)2, then either all these numbers are nonnegative or all these numbers are nonpositive.

Solution

Let us suppose that c1 \leqc2 \leq\cdot \cdot \cdot \leqcn and that c1 < 0 < cn. There exists k, 1 \leqk < n, such that ck \leq0 < ck+1. Then we have (n −1)(c2 1 + c2 2 + \cdot \cdot \cdot + c2 n) \geqk(c2 1 + \cdot \cdot \cdot + c2 k) + (n −k)(c2 k+1 + \cdot \cdot \cdot + c2 n) \geq(c1 + \cdot \cdot \cdot + ck)2 + (ck+1 + \cdot \cdot \cdot + cn)2 = (c1 + \cdot \cdot \cdot + cn)2 −2(c1 + \cdot \cdot \cdot + ck)(ck+1 + \cdot \cdot \cdot + cn), from which we obtain (c1 + \cdot \cdot \cdot + ck)(ck+1 + \cdot \cdot \cdot+ cn) \geq0, a contradiction. Second solution. By the given condition and the inequality between arith- metic and quadratic mean we have (c1 + \cdot \cdot \cdot + cn)2 = (n −1)(c2 1 + \cdot \cdot \cdot + c2 n−1) + (n −1)c2 n \geq(c1 + \cdot \cdot \cdot + cn−1)2 + (n −1)c2 n, which is equivalent to 2(c1 + c2 + \cdot \cdot \cdot + cn)cn \geqnc2 n. Similarly, 2(c1 + c2 + \cdot \cdot \cdot + cn)ci \geqnc2 i for all i = 1, . . . , n. Hence all ci are of the same sign.