Let $x_1,x_2,...,x_n$ be real numbers and A be the average of those numbers. Prove that $x_i\ge A$ for some $i$.

Question

Let $x_1,x_2,...,x_n$ be real numbers and A be the average of those numbers. Prove that $x_i\ge A$ for some $i$.

I understand this intuitively, but I can't seem to figure out how to prove it. I know it's supposed to be an inequality argument where we assume $x_i < A$ for all $i$ and prove by contradiction, but I can't get any further than that.

Answer 1

You're right that it's going to be a proof by contradiction:

Suppose that for each $i$, we have $x_i<A$. Then $x_1+x_2+ \cdots + x_n<A+A+\cdots+A=nA$, and dividing both sides by $n$ yields $ \dfrac {x_1+x_2+ \cdots + x_n}{n}<A$. But this is a contradiction.

Answer 2

It does not have to be by contradiction if you rephrase it slightly:

If $x_k < L$ for all $k$ then ${1 \over n} \sum_k x_k < L$. Since ${1 \over n} \sum_k x_k = A$ we must have $x_k \ge A$ for some $k$.