### 34th IMO 1993 shortlist

**Problem 24**
Show that a/(b + 2c + 3d) + b/(c + 2d + 3a) + c/(d + 2a + 3b) + d/(a + 2b + 3c) ≥ 2/3 for any positive reals.

**Solution**

Put A = b+2c+3d, B = c+2d+3a, C = d+2a+3b, D = a+2b+3c. Then a = (-5A+7B+C+D)/24, b = (-5B+7C+D+A), etc. So lhs = -20/24 + (7B/A + C/A + D/A + 7C/B + D/B + A/B + 7D/C + A/C + B/C + 7A/D + B/D + C/D)/24. We can regard 7B/A as B/A + B/A + ... + B/A, so the bracket has 36 terms with product 1. Hence by AM/GM it is ≥ 36. So we get lhs ≥ -20/24 + 36/24 = 2/3.

34th IMO shortlist 1993

© John Scholes

jscholes@kalva.demon.co.uk

8 Nov 2003

Last corrected/updated 8 Nov 03