8. If $f$ is a continuous real function such that $f(x-1)+f(x+1) \geq x+f(x)$ for all $x$, what is the minimum possible value of $\int_{1}^{2005} f(x) d x$ ?
Solution: 2010012
Let $g(x)=f(x)-x$. Then
$$
g(x-1)+x-1+g(x+1)+x+1 \geq x+g(x)+x,
$$
or $g(x-1)+g(x+1) \geq g(x)$. But now,
$$
g(x+3) \geq g(x+2)-g(x+1) \geq-g(x) .
$$

Therefore
$$
\begin{aligned}
\int_{a}^{a+6} g(x) d x & =\int_{a}^{a+3} g(x) d x+\int_{a+3}^{a+6} g(x) d x \\
& =\int_{a}^{a+3}(g(x)+g(x+3)) d x \geq 0 .
\end{aligned}
$$

It follows that
$$
\int_{1}^{2005} g(x)=\sum_{n=0}^{333} \int_{6 n+1}^{6 n+7} g(x) d x \geq 0,
$$
so that
$$
\int_{1}^{2005} f(x) d x=\int_{1}^{2005}(g(x)+x) d x \geq \int_{1}^{2005} x d x=\left[\frac{x^{2}}{2}\right]_{1}^{2005}=\frac{2005^{2}-1}{2}=2010012 .
$$

Equality holds for $f(x)=x$.