IMO 1989 SL 26

Origin: KOR

Problem

Let n be a positive integer and let a, b be given real numbers. Determine the range of x0 for which n  i=0 xi = a and n  i=0 x2 i = b, where x0, x1, . . . , xn are real variables.

Solution

By the Cauchy–Schwarz inequality,

n  i=1 xi \leqn n  i=1 x2 i . Since n i=1 xi = a −x0 and n i=1 x2 i = b −x2 0, we have (a −x0)2 \leq n(b −x2 0), i.e., (n + 1)x2 0 −2ax0 + (a2 −nb) \leq0. The discriminant of this quadratic is D = 4n(n + 1) / b −a2/(n + 1) , so we conclude that (i) if a2 > (n + 1)b, then such an x0 does not exist; (ii) if a2 = (n + 1)b, then x0 = a/n + 1; and

(iii) if a2 < (n + 1)b, then a− \sqrt D/2 n+1 \leqx0 \leqa+ \sqrt D/2 n+1 . It is easy to see that these conditions for x0 are also sufficient.