# 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 $i$.

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.

• Can you confirm average to indicate arithmetic mean? Commented Oct 22, 2017 at 22:18

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.

• ah i understand now. thank you very much.
– d.v.
Commented Oct 22, 2017 at 22:17
• @d.v. You're welcome!
– Ovi
Commented Oct 22, 2017 at 22:18
• It may be worth noting that this proof can be generalised to any mean relating to a function(see generalised f mean) as the function has to be convex. Commented Oct 22, 2017 at 22:22
• +1! But your answer will be better if you replace contradiction by contraposition. Just assume x's are all less than K, for some K, and conclude that A<K too. Commented Oct 22, 2017 at 22:31
• @user334639 Thank you, yes that would indeed be better but I think I will just leave the answer as it is since you already gave a sketch for how to use contraposition.
– Ovi
Commented Oct 22, 2017 at 23:04

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 $k$.

• That looks like contradiction to me. Commented Oct 22, 2017 at 23:26
• @Qudit: It would be the contrapositive, not a contradiction. Commented Oct 22, 2017 at 23:28
• Restructuring it in this way changes nothing about the argument. It just hides the contradiction inside an implication. It's only different from the idea of proof by contradiction if you define contradiction in a narrow technical sense. Commented Oct 22, 2017 at 23:30
• It is so different from contradiction! The answer never assumed the opposite of what it intends to conclude. It's not hiding a contradiction anywhere. Commented Oct 22, 2017 at 23:34
• Frankly, I find proof by contradiction to be unsatisfactory, while I am happy to accept the tautological equivalence of $P \rightarrow Q$ and $\lnot Q\rightarrow \lnot P$ without qualm. This is subjective, of course, but I find contradictions destructive in some sense :-). Commented Oct 22, 2017 at 23:36