Part of sum inequality to number of terms Suppose you have a series
 $$ S=\sum_{i=1}^n a_i$$ and rearrange the terms such that, $$ a_1\geq a_2 \geq \cdots \geq a_l \geq a_i\geq 0 $$ for all $$ i > l$$, then it should be obvious that $$ \frac{a_1+a_2+\cdots+a_l}{S}\geq \frac{l}{n}$$ How would one go about proving this? Using upper and underestimates won't work.

Hint: first rearrange your inequality to $\displaystyle \frac{S}{n} \leqslant \frac{a_1+a_2+\ldots+a_l}{l}$. Then note that
$$\frac{S}{n} = \frac{l}{n} \cdot \frac{a_1+a_2+\ldots+a_l}{l} + \left( 1-\frac{l}{n} \right) \cdot \frac{a_{l+1} + \ldots + a_n}{n-l}$$
which is a weighted arithmetic mean of two numbers, where the first $\frac{a_1+\ldots+a_l}{l}$ is bigger.
