Proof: By the AM-GM inequality,
$$x^{2009}-2008(x-1)=x^{2009}+1+\ldots \ldots .+1_{(2008 \uparrow 1)}-2008 x \geq 2009 x-2008 x=x,$$

Therefore, $\frac{x^{2009}-2008(x-1)}{y+z} \geq \frac{x}{y+z}$,
Similarly, $\frac{y^{2009}-2008(y-1)}{z+x} \geq \frac{y}{z+x}, \frac{z^{2009}-2008(z-1)}{x+y} \geq \frac{z}{x+y}$,
By the Cauchy-Schwarz inequality,
$$\begin{array}{l} \text { Hence } \frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y} \\ \geq \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{x^{2}}{x(y+z)}+\frac{y^{2}}{y(z+x)}+\frac{z^{2}}{z(x+y)} \\ \geq \frac{(x+y+z)^{2}}{x(y+z)+y(z+x)+z(x+y)}=\frac{(x+y+z)^{2}}{2(x y+y z+z x)} \geq \frac{3(x y+y z+z x)}{2(x y+y z+z x)}=\frac{3}{2} . \\ \text { And } \frac{1}{2}(x+y+z) \leq \frac{1}{2} \sqrt{3\left(x^{2}+y^{2}+z^{2}\right)}=\frac{1}{2} \sqrt{3 \times 3}=\frac{3}{2} . \\ \text { Therefore, } \frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y} \geq \frac{1}{2}(x+y+z) . \end{array}$$