**The question is below.
Let $(s_n)$ be a nondecreasing sequence of positive numbers and define $\sigma_n=\frac{1}{n}(s_1+s_2+\cdots+s_n)$. Prove that $(\sigma_n)$ is a nondecreasing sequence.
**In my text book, the monotone sequence is defined as below.
A sequence$(s_n)$ of real numbers is called a nondecreasing sequence if $s_n \leq s_{n+1}, \forall n$, and nonincreasing sequence if $s_n \geq s_{n+1}, \forall n$.
**My work is as follows.
$$\sigma_1 = \frac{s_1}{1} \Rightarrow \sigma_1 = s_1$$
$$\sigma_2 = \frac{s_1+s_2}{2}\Rightarrow 2*\sigma_2 = s_1+s_2$$
$$\sigma_3 = \frac{s_1+s_2+s_3}{3}\Rightarrow 3*\sigma_3 = s_1+s_2+s_3$$
$$\vdots$$
$$\sigma_n = \frac{s_1+s_2+\cdots+s_n}{n}\Rightarrow n*\sigma_n = s_1+s_2+\cdots+s_n$$
Since $(s_n)$ is nondecreasing sequence of positive numbers and $1*\sigma_1 \leq 2*\sigma_2 \leq \cdots \leq n*\sigma_n$, $\sigma_n$ is nondecreasing sequence. Q.E.D.
I am afraid is could be wrong because I proved it with coefficient for each sequence $\sigma_n$. Can anyone give me some hints if this is wrong??
Best Answer
Suppose that $$\frac1{n+1}(s_1+\ldots+s_n+s_{n+1})<\frac1n(s_1+\ldots+s_n)\;;$$ then
$$\begin{align*}s_1+\ldots+s_n+s_{n+1}&<\frac{n+1}n(s_1+\ldots+s_n)\\ &=\left(1+\frac1n\right)(s_1+\ldots+s_n)\\ &=(s_1+\ldots+s_n)+\frac1n(s_1+\ldots+s_n)\;, \end{align*}$$
and therefore $$s_{n+1}<\frac1n(s_1+\ldots+s_n)\;.$$ But then $$ns_{n+1}<s_1+\ldots+s_n\le ns_n\;,$$ and $s_{n+1}<s_n$, which is impossible.