If the average of numbers is $N$, prove that at least one of them is $\ge N$

Question

There is a set a set $S$ of numbers. i.e. $(s_1, s_2, s_3, s_4, s_5, ..., s_n)$. The average of the numbers in the set is $N$. How do I prove that at least one of the numbers in the set is greater than or equal to $N$?

I am thinking that a proof by contradiction is probably the easiest way.

I could do a proof by contradiction by supposing that no $s_i$ in $S$ is greater than or equal to $N$. But then I am stuck on how to proceed to show that it is False and the original statement is true.

Answer 1

Suppose there are $n$ elements $s_i \in S$ and the largest is $M$.

Then looking at the mean we have $\dfrac{\sum_i s_i} {n}=N$, and each $s_i \le M$,

so $ nN=\sum_i s_i \le \sum_i M=nM$ meaning $nN \le nM$ and so $$N \le M$$ i.e. the mean is less than or equal to the maximum.

No need to look for a contradiction.

Answer 2

Enumerate the elements of $S$ as $S=\{s_1,\ldots,s_K\}$. Suppose that it was true that $s_i<N$ for all $i$. Then, $$ \sum_{i=1}^K s_i<\sum_{i=1}^K N=K\times N=K\frac{\sum_{i=1}^K s_i}{K}=\sum_{i=1}^K s_i. $$ The leftmost term and rightmost term are the same so you have a contradiction.