IMO 1997 SL 19

Origin: IRE

Problem

Let a1 \geq\cdot \cdot \cdot \geqan \geqan+1 = 0 be a sequence of real numbers. Prove that n  k=1 ak \leq n  k=1 \sqrt k(\sqrtak −\sqrtak+1).

Solution

Using that an+1 = 0 we can transform the desired inequality into

a1 + a2 + \cdot \cdot \cdot + an+1 \leq \sqrt 1\sqrta1 + ( \sqrt 2 − \sqrt 1)\sqrta2 + \cdot \cdot \cdot + ( \sqrt n + 1 −\sqrtn)\sqrtan+1. (1) We shall prove by induction on n that (1) holds for any a1 \geqa2 \geq\cdot \cdot \cdot \geq an+1 \geq0, i.e., not only when an+1 = 0. For n = 0 the inequality is

obvious. For the inductive step from n −1 to n, where n \geq1, we need to prove the inequality

a1 + \cdot \cdot \cdot + an+1 −\sqrta1 + \cdot \cdot \cdot + an \leq( \sqrt n + 1 −\sqrtn)\sqrtan+1. (2) Putting S = a1 + a2 + \cdot \cdot \cdot + an, this simplifies to

S + an+1 − \sqrt S \leq \sqrtnan+1 + an+1 −\sqrtnan+1. For an+1 = 0 the inequality is obvious. For an+1 > 0 we have that the function f(x) = \sqrtx + an+1 −\sqrtx = an+1 \sqrtx+an+1+\sqrtx is strictly decreasing on R+; hence (2) will follow if we show that S \geqnan+1. However, this last is true because a1, . . . , an \geqan+1. Equality holds if and only if a1 = a2 = \cdot \cdot \cdot = ak and ak+1 = \cdot \cdot \cdot = an+1 = 0 for some k. Second solution. Setting bk = \sqrtak −\sqrtak+1 for k = 1, . . . , n we have ai = (bi + \cdot \cdot \cdot + bn)2, so the desired inequality after squaring becomes n  k=1 kb2 k + 2  1\leqk<l\leqn kbkbl \leq n  k=1 k b2 k + 2  1\leqk<l\leqn \sqrt kl bkbl, which clearly holds.