a, b, c, d are non-negative reals with sum 1. Show that abc + bcd + cda + dab ≤ 1/27 + 176abcd/27.
Solution
Put f(a,b,c,d) = abc + bcd + cda + dab - 176abcd/27. Note that the order of the arguments is unimportant. We can write f(a,b,c,d) = ab(c+d) + cd(a+b-176ab/27). Suppose the result is false. If a+b-176ab/27 ≤ 0, then f(a,b,c,d) ≤ ab(c+d) ≤ 1/27 (using AM/GM for a,b,c+d). Contradiction. So we must have a+b-176ab/27 > 0. Hence by AM/GM applied to c,d we get f(a,b,c,d) ≤ ab(c+d) + ¼(c+d)2(a+b-176ab/27) = f(a,b,(c+d)/2,(c+d)/2).
We now repeat the argument, getting successively f(a,b,c,d) ≤ f(a,b,(c+d)/2,(c+d)/2) ≤ f((a+b)/2,(a+b)/2,(c+d)/2,(c+d)/2) ≤ f(¼,¼,(a+b)/2,(c+d)/2) ≤ f(¼¼,¼,¼) = 1/27. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
8 Nov 2003
Last corrected/updated 8 Nov 03