a and b are positive integers and c = (aa+1 + bb+1)/(aa + bb). By considering (xn - nn)/(x - n) or otherwise, show that ca + cb ≥ aa + bb.
Solution
(xn - nn)/(x - n) = xn-1 + xn-2n + ... + nn-1. If x > n, then each term is > nn-1, so (xn - nn)/(x - n) > nn. If x < n, then each term is <= nn with equality only for the last term, so (xn - nn)/(x - n) < nn. So multiplying by x - n, which is positive for x > n and negative for x < n, we get (xn - nn) > (x - n)nn except for x = n, and hence (xn - nn) ≥ (x - n)nn for all x ≥ 0.
Putting x = c, n = a, we get (ca - aa) ≥ (c - a)aa. Similarly, putting x = c, n = b, we get (cb - bb) ≥ (c - b)bb. Adding (ca + cb) - (aa + bb) ≥ (c - a)aa + (c - b)bb = c(aa + bb) - aa+1 - bb+1 = 0, which is the required result.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002