A sequence xn of positive reals satisfies xn-1xn+1 ≤ xn2. Let an be the average of the terms x0, x1, ... , xn and bn be the average of the terms x1, x2, ... , xn. Show that anbn-1 ≥ an-1bn.
Solution
Put k = x1 + x2 + ... + xn-1. We have to show that (x0 + k + xn)/(n+1) k/(n-1) ≥ (x0 + k)/n (k + xn)/n or n2(k + x0 + xn) ≥ (n2 - 1)(k + x0)(k + xn). Simplifying slightly, this is equivalent to k(k + x0 + xn) ≥ (n2 - 1)x0xn. So it is evidently sufficient to show that k ≥ (n-1)(x0xn)1/2. [Then AM/GM gives x0 + xn ≥ 2(x0xn)1/2, so (k + x0 + xn) ≥ (n+1)(x0xn)1/2.]
We have x0/x1 ≤ x1/x2 ≤ ... ≤ xn-1/xn. Hence (apparently weakening) x0xn ≤ x1xn-1 ≤ x2xn-2 ≤ ... . So k = (x1 + xn-1) + (x2 + xn-2) + ... ≥ (n-1) (x0xn)1/2, using first AM/GM, then the relation just established.
© John Scholes
jscholes@kalva.demon.co.uk
23 Aug 2002