parmenides51 wrote: Assume : $x,y,z>0$ , $ x^2+y^2+z^2 = 3 $ . Prove the following inequality : $${\frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{x+z}+\frac{z^{2009}-2008(z-1)}{x+y}\ge\frac{1}{2}(x+y+z)}$$ https://artofproblemsolving.com/community/c6h608420p3616030

By Bernoulli $x^{2009}=(1+(x-1))^{2009}\ge 1+2009(x-1)$ hence it suffices to prove $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{1}{2}(x+y+z).$$By C-S it suffices to prove $$\frac{(x+y+z)^2}{xy+yz+zx}\ge x+y+z$$or $$\left(x^2+y^2+z^2\right)(x+y+z)^2\ge 3(xy+yz+zx)^2$$which is obviously true.

By AM-GM and Nesbit $\sum\frac{x^{2009}-2008x+2008}{y+z}\geq\sum\frac{x}{y+z}\geq\frac{3}{2}=\frac{\sqrt{3(x^2+y^2+z^2)}}{2}\geq\frac{x+y+z}{2}$.