x1, ... , xn are positive reals. sk is the sum of all products of k of the xi (for example, if n = 3, s1 = x1 + x2 + x3, s2 = x1x2 + x2x3 + x3x1, s3 = x1x2x3). Show that sksn-k ≥ (nCk)2 sn for 0 < k < n.
Solution
Each of sk and sn-khave nCk terms. So we may multiply out the product sksn-k to get a sum of (nCk)2 terms. We now apply the arithmetic/geometric mean result. The product of all the terms must be a power of sn by symmetry and hence must be sn to the power of (nCk)2. So the geometric mean of the terms is just sn. Hence result.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002