Show that $\min\{a_1,a_2,...,a_n\}$ is maximum when $a_1=a_2=...=a_n$.

Question

Given $a_1,a_2,...,a_n\in\mathbb R$, and $a_1+a_2+...+a_n=A$. Show that $\min\{a_1,a_2,...,a_n\}$ is maximum when $a_1=a_2=...=a_n$.

I feel this is quite a common sense but I don't know how to prove it. Thanks.

Answer 1

Assume, without loss of generality, that $a_1\leq a_2\leq\dots\leq a_n$. Now, $$A=\sum_{i=1}^n a_i\geq na_1=n\min\{a_1,\dots,a_n\}.$$ Hence $\min\{a_1,\dots,a_n\}\leq \min\{A/n,\dots,A/n\}$ as desired.

Answer 2

Suppose that $\min\{a_1,...,a_n\}$ is maximum for some other sequence, call the minimum element in the set $a_m$. Then since $a_1+\cdots a_n=A$, $na_m< A\implies a_m< \frac{A}{n}$, but the constant sequence $a_1,...,a_n=\frac{A}{n}$ satisfies $a_1+\cdots a_n=A$, hence we have a contradiction.