22nd ASU 1988 Problem 15 What is the minimal value of b/(c + d) + c/(a + b) for positive real numbers b and c and non-negative real numbers a and d such that b + c ≥ a + d? Solution
: √2 - 1/2. Obviously a + d = b + c at the minimum value, because increasing a or d reduces the value. So we may take d = b + c - a. We also take b >= c (interchanging b and c if necessary). Dividing through by b/2 shows that there is no loss of generality in taking b = 2, so 0 = 0, so c 2 + c(2k - 2h) + h 2 - 2hk + k 2 >= 0. Hence c 2 + ck + h 2 >= (2h - k)(c + k). Hence c/h 2 + 1/(c+k) >= 2/h - k/h 2 with equality iff c = h - k. Applying this to c/(a+2) + 1/(c+1-a/2) where h = √(a+2), k = 1-a/2, we find that c/(a+2) + 2/(2c+2-a) >= 2/√(a+2) + (a-2)/(2a+4). The allowed range for c is 0 ≤ c ≤ 2 and 0 ≤ a ≤ c+2, hence 0 ≤ a ≤ 4. Put x = 1/√(a+2), so 1/√6 <= x <= 1/√2. Then 2/√(a+2) + (a-2)/(2a+4) = 2x + 1/2 - 2x 2 = 1 - (2x-1) 2 /2. We have -0.184 = (2/√6 - 1) ≤ 2x- 1 ≤ √2 - 1 = 0.414. Hence c/(a+2) + 2/(2c+2-a) ≥ 1 - (√2 - 1) 2 /2 = √2 - 1/2. We can easily check that the minimum is achieved at b = 2, c = √2 - 1, a = 0, d = √2 + 1. 22nd ASU 1988 © John Scholes jscholes@kalva.demon.co.uk 26 June 2002