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.

Note:
$$\frac{a_1+\cdots +a_l}{S}\ge \frac{l}{n} \Rightarrow $$
$$\frac{a_1+\cdots+a_l}{l}\ge \frac{a_1+\cdots +a_l+a_{l+1}+\cdots +a_n}{l+(n-l)} \Rightarrow$$
$$(n-l)(a_1+\cdots +a_l)\ge l(a_{l+1}+\cdots +a_n)$$
Now the LHS:
$$(n-l)(a_1+\cdots +a_l)\ge (n-l)(\underbrace{a_l+\cdots +a_l}_{l})$$
and the RHS:
$$l(\underbrace{a_{l+1}+\cdots a_{l+1}}_{n-l})\ge l(a_{l+1}+\cdots +a_n)$$
Hence:
$$(n-l)la_l\ge l(n-l)a_{l+1} \Rightarrow$$
$$a_l\ge a_{l+1}.$$
