Find all solutions in real numbers x1, x2, ... , xn+1 all at least 1 such that: (1) x11/2 + x21/3 + x31/4 + ... + xn1/(n+1) = n xn+11/2; and (2) (x1 + x2 + ... + xn)/n = xn+1.
Answer
The only solution is the obvious, all xi = 1.
Solution
By Cauchy-Schwartz, (∑ xi1/2)2 ≤ (∑ 1)(&sum xi), with equality iff all xi equal. In other words, if we put xn+1 = (x1 + x2 + ... + xn)/n, then ∑ xi1/2 ≤ n xn+11/2. But since all xi ≥ 1, we have x11/2 + x21/3 + x31/4 + ... + xn1/(n+1) ≤ ∑ xi1/2 with equality iff x2 = x3 = ... = xn = 1. Hence x11/2 + x21/3 + x31/4 + ... + xn1/(n+1) ≤ xn+11/2 with equality iff all xi = 1.
© John Scholes
jscholes@kalva.demon.co.uk
22 January 2004
Last corrected/updated 22 Jan 04