Extended Pigeonhole Principle: How to prove it?

Question

A version of the pigeonhole principle is:

(1) If m objects are put in n boxes and n < m, then at least one box contains at least ceil(m/n) objects

An alternate (more generalized) version is:

(2) For a nonempty finite collection of integers (not necessarily distinct), the maximum value is at least the average value.

How do I prove (2) starting from (1)? Every description I have read leaves this as obvious but I can't seem to make the link so easily.

Answer 1

Assume you have $n$ integers $a_1,\dots,a_n$. Assign them $n$ "boxes" $A_1,\dots,A_n$, such that for all $i$, $A_i$ is filled with $a_i$ "objects". The total number of objects will be $m:=a_1+\dots+a_n$. (1) applies and tells, that there is a box $A_i$ which contains at least $ceil(m/n)$ objects. Now $a_i$ is the number of objects of $A_i$, so: $$a_i\geq ceil(m/n)\geq m/n = \frac{a_1+\dots+a_n}{n}$$

Answer 2

I don't know how to prove (2) from (1), but (2) can be proved quite easily, can't it?

Let $a_1,\dots,a_n\in\mathbb R$ (not even necessarily in $\mathbb{Z}$). Suppose for contradiction that $$a_j< \frac 1n \sum_{i} a_i \quad\text{for all }j$$ If you add all these $n$ inequalities, you get $$\sum_j{a_j} < \sum_{i} a_i,$$ contradiction (since the name of the summation variable plays no role).

Do I miss something in your question?