64th Putnam 2003

a₁, a₂, ... , a_n, b₁, ... , b_n are non-negative reals. Show that (∏ a_i)^1/n + (∏b_i)^1/n ≤ (∏(a_i+b_i))^1/n.

Solution

If any a_i or b_i is zero, then the result is trivial, so assume all are positive. By AM/GM (∏a_i/(a_i+b_i))^1/n ≤ (∑a_i/(a_i+b_i))/n. Similarly for b_i. Add.

64th Putnam 2003

© John Scholes
jscholes@kalva.demon.co.uk
8 Dec 2003
Last corrected/updated 8 Dec 03