Russian 2002

Show that √x + √y + √z ≥ xy + yz + zx for positive reals x, y, z with sum 3.

Solution

x² + √x + √x ≥ 3x by AM/GM. Adding similar inequalities for y, z, we get x² + y² + z² + 2(√x + √y + √z) ≥ 3(x + y + z) = (x + y + z)² = x² + y² + z² + 2(xy + yz + zx).

Thanks to Suat Namli

Russian 2002

© John Scholes
jscholes@kalva.demon.co.uk
31 December 2003
Last corrected/updated 31 Dec 03