x1, x2, ... , xn are reals not less than 1. Show that 1/(1 + x1) + 1/(1 + x2) + ... + 1/(1 + xn) ≥ n/(1 + (x1x2 ... xn)1/n).
Solution
For n = 2 we have to show that 1/(1 + x) + 1/(1 + y) ≥ 2/(1 + (xy)1/2) or 2( (xy)1/2 - 1)( (x+y)/2 - (xy)1/2) ≥ 0. The first factor is non-negative since x and y ≥ 1 and the second is non-negative by the AM/GM.
If it is true for n, then the result for 2 means it is also true for 2n, because we can apply the result for 2 to 1/(1 + (x1x2 ... xn)1/n) + 1/(1 + (xn+1xn+2 ... x2n)1/n). So by induction it holds for n a power of 2. But that means it hold for any n, because we just take the result for 2m ≥ n and then set the surplus terms equal to (x1...xn)1/n.
© John Scholes
jscholes@kalva.demon.co.uk
30 Aug 2002