MathDB
sum (x^{2009}-2008(x-1))/(y+z) >= 1/2 (x+y+z), if x,y,z>0, x^2+y^2+z^2 = 3

Source: China Northern MO 2009 p5 CNMO

December 12, 2020
algebrainequalities

Problem Statement

Assume : x,y,z>0x,y,z>0 , x2+y2+z2=3 x^2+y^2+z^2 = 3 . Prove the following inequality : x20092008(x1)y+z+y20092008(y1)x+z+z20092008(z1)x+y12(x+y+z){\frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{x+z}+\frac{z^{2009}-2008(z-1)}{x+y}\ge\frac{1}{2}(x+y+z)}