Let $ \Delta f(x,y) = \Delta f(y,x) = \frac {f(x) - f(y)}{x - y}$. Since $ f(x)$ is a convex function, then $ \Delta f(x,y)$ is an increasing function for the both variables. $ \sum_{k = 1}^{2003}f(a_{k + 1})a_k - \sum_{k = 2}^{2004}f(a_{k - 1})a_{k} =$ $ \sum_{k = 1}^{2003}a_k(a_{k + 1} - a_{k - 1})\Delta f(a_{k + 1},a_{k - 1})$ $ = \sum_{k = 1}^{2003}a_ka_{k + 1}\Delta f(a_{k + 1},a_{k - 1}) - \sum_{k = 1}^{2003}a_ka_{k + 1}\Delta f(a_{k + 2},a_{k}) =$ $ \sum_{k = 1}^{2003}a_ka_{k + 1}\left[\Delta f(a_{k + 1},a_{k - 1}) - \Delta f(a_{k + 2},a_{k})\right]\ge 0$, since $ a_i\ge a_{i + 1}$. (That's not the complete solution, but the rest is easy )