IMO 2001 SL A3

Origin: ROM | Category: Algebra

Problem

Let x1, x2, . . . , xn be arbitrary real numbers. Prove the inequality x1 1 + x2 + x2 1 + x2 1 + x2

• \cdot \cdot \cdot + xn 1 + x2 1 + \cdot \cdot \cdot + x2n < \sqrtn.

Solution

By the arithmetic–quadratic mean inequality, it suffices to prove that x2 (1 + x2 1)2 + x2 (1 + x2 1 + x2 2)2 + \cdot \cdot \cdot + x2 n (1 + x2 1 + \cdot \cdot \cdot + x2n)2 < 1. Observe that for k \geq2 the following holds: x2 k (1 + x2 1 + \cdot \cdot \cdot + x2 k)2 \leq x2 k (1 + \cdot \cdot \cdot + x2 k−1)(1 + \cdot \cdot \cdot + x2 k)

1 + x2 1 + \cdot \cdot \cdot + x2 k−1 − 1 + x2 1 + \cdot \cdot \cdot + x2 k . For k = 1 we have x2 (1+x1)2 \leq1 − 1+x2 1 . Summing these inequalities, we obtain x2 (1 + x2 1)2 + \cdot \cdot \cdot + x2 n (1 + x2 1 + \cdot \cdot \cdot + x2n)2 \leq1 − 1 + x2 1 + \cdot \cdot \cdot + x2n < 1. Second solution. Let an(k) = sup  x1 k2+x2 1 + \cdot \cdot \cdot + xn k2+x2 1+\cdot\cdot\cdot+x2n

and an = an(1). We must show that an < \sqrtn. Replacing xi by kxi shows that an(k) = an/k. Hence an = sup x1

x1 1 + x2 + an−1

1 + x2

= sup \theta (sin \theta cos \theta + an−1 cos \theta), (1) where tan \theta = x1. The above supremum can be computed explicitly: an = \sqrt  3an−1 +

a2 n−1 + 8   4 −a2 n−1 + an−1

a2 n−1 + 8. However, the required inequality is weaker and can be proved more easily: if an−1 < \sqrtn −1, then by (1) an < sin \theta+\sqrtn −1 cos \theta = \sqrtn sin(\theta+\alpha) \leq \sqrtn, for \alpha \in(0, \pi/2) with tan \alpha = \sqrtn.