b > 0 and a are fixed real numbers and n is a fixed positive integer. The real numbers x0, x1, ... , xn satisfy x0 + x1 + ... + xn = a and x02 + ... + xn2 = b. Find the range of x0.
Solution
By Cauchy we have (∑ xi)2 ≤ n ∑ xi2 (for n terms) (*). Hence if x0 = x, then (a - x)2 ≤ n(b - x2) or (n+1) x2 - 2ax + a2 - nb ≤ 0.
If a2 < (n+1)(a2 - nb), or (n+1)b < a2, then the quadratic is always positive, so there are no possible values for x0. If a2 = (n+1)b, then the only possible value of x0 is a/(n+1). If a2 < (n+1)b, then x0 must lie between a/(n+1) + (1/(n+1) )√( n(n+1)b - na2) and a/(n+1) - (1/(n+1) )√( n(n+1)b - na2).
We get equality in the Cauchy inequality (*) iff all the terms xi are equal. So all the values in the range given above can be achieved by taking xi = (a - x0)/n for i > 0.
© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02