26th USAMO 1997

Show that xyz/(x³ + y³ + xyz) + xyz/(y³ + z³ + xyz) + xyz/(z³ + x³ + xyz) ≤ 1 for all real positive x, y, z.

Solution

For positive x, y, x ≥ y iff x² ≥ y², so (x - y)(x² - y²) ≥ 0, or x³ + y³ ≥ xy(x + y). Hence x³ + y³ + xyz ≥ xy(x + y + z) and so xyz/(x³ + y³ + xyz) ≤ z/(x + y + z). Adding the two similar equations gives the required inequality.

26th USAMO 1997

© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002