Theorem.
If the sequence $\{a_n\}$ is increasing, then their arithmetic mean \[
\sigma_n:=\frac{a_1+\ldots+a_n}{n} \] is also increasing.
\begin{align*}
\sigma_{n+1}-\sigma_n &=\frac{a_1+\ldots+a_{n+1}}{n+1}-\frac{a_1+\ldots+a_n}{n} \\
&=\frac{1}{n+1}a_{n+1}-\frac{1}{n(n+1)}(a_1+\ldots+a_n)\\
&=\frac{na_{n+1}-(a_1+\ldots+a_n)}{n(n+1)}\\
&=\frac{(a_{n+1}-a_1)+\ldots+(a_{n+1}-a_n)}{n(n+1)}\geq 0. \qquad\blacksquare \end{align*}
\sigma_{n+1}-\sigma_n &=\frac{a_1+\ldots+a_{n+1}}{n+1}-\frac{a_1+\ldots+a_n}{n} \\
&=\frac{1}{n+1}a_{n+1}-\frac{1}{n(n+1)}(a_1+\ldots+a_n)\\
&=\frac{na_{n+1}-(a_1+\ldots+a_n)}{n(n+1)}\\
&=\frac{(a_{n+1}-a_1)+\ldots+(a_{n+1}-a_n)}{n(n+1)}\geq 0. \qquad\blacksquare \end{align*}
