Average Proofs, where $n\ge 2$ is greater than the average of real numbers

Question

Prove that at least one of the real numbers $a_1; a_2;... a_n,$ where $n \ge 2$, is greater than or equal to the average of these numbers.

I'm stuck on how I am supposed to prove this. Honestly proofs are my worst skill in all of my math classes.

Answer 1

Suppose that all $a_i$ are less than the average, which we'll call $a$, so $a_i<a$. Then $$ a=\frac1n\sum_{i=1}^n a_i<\frac1n\sum_{i=1}^n a=a. $$ This is a contradiction.

Answer 2

Without loss of generality, we can suppose that $a_1\le a_2\le\dots\le a_n$ (if not, we can just reindex them). If they are all less than the average, then in particular,$$a_n<\frac1n\sum_{k=1}^na_k\le\frac1n\sum_{k=1}^na_n=\frac{na_n}n=a_n.$$