n > 1, and a and b are positive real numbers such that an - a - 1 = 0 and b2n - b - 3a = 0. Which is larger?
Solution
Answer: a > b.
Note that an = a + 1 > 1 (since a is positive). Hence a > 1. So a2n = (a + 1)2 = a2 + 2a + 1. Put a = 1 + k, then a2 = 1 + 2k + k2 > 1 + 2k, so a2 + 1 > 2 + 2k = 2a. Hence a2n > 4a. So (b/a)2n = (b + 3a)/a2n < (b + 3a)/4a. If b/a ≥ 1, then (b + 3a)/4a ≤ (b + 3b)/4a = b/a, so (b/a)2n < b/a. Contradiction. Hence b/a < 1.
© John Scholes
jscholes@kalva.demon.co.uk
23 Aug 2002