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.
Best Answer
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.