IMO 1972 SL 3

Let x1, x2, . . . , xn be real numbers satisfying x1+x2+\cdot \cdot \cdot+xn =

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IMO 1972 SL 3

Origin: CZS

Problem

Let x1, x2, . . . , xn be real numbers satisfying x1+x2+\cdot \cdot \cdot+xn = 0. Let m be the least and M the greatest among them. Prove that x2 1 + x2 2 + \cdot \cdot \cdot + x2 n \leq−nmM.

Solution

We have for each k = 1, 2, . . . , n that m \leqxk \leqM, which gives (M − xk)(m −xk) \leq0. It follows directly that 0 \geq n  k=1 (M −xk)(m −xk) = nmM −(m + M) n  k=1 xk + n  k=1 x2 k. But n k=1 xk = 0, implying the required inequality.