Prove that (n+1)mnm ≥ (n+m)!/(n-m)! ≥ 2mm! for all positive integers n, m with n ≥ m.
Solution
For any integer k ≥ 1, we have (n + k)(n - k + 1) = n2 + n - k2 + k ≤ n(n + 1). Taking the product from k = 1 to m we get (n + m)!/(n - m)! ≤ (n + 1)mnm.
For k = 1, 2, ... , m, we have n ≥ k and hence n + k ≥ 2k. Taking the product from k = 1 to m, we get (n + m)!/(n - m)! ≥ 2mm! .
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002