Prove that $\liminf_{n→∞} s_n \le \liminf_{n→∞} σ_n$ 
I am trying to prove that $$\liminf_{n→∞}  s_n \le \liminf_{n→∞} σ_n$$
  given that $σ_n=\frac1n(s_1+s_2+\dots+s_n)$.   

Setting $\alpha = \liminf_{n→∞}s_n$, hence $$\forall \epsilon>0 \ \exists N: \forall n \geq N, \ \alpha - \epsilon <\inf s_n \le s_n$$ Now, I know I can split $σ_n$ to get $\sigma_n \geq \frac{1}{n}\ (s_1 +\dots + s_N) + \frac{1}{n}\ (s_{N+1}+\dots+s_n)$ and later\begin{align}\sigma_n &\geq \frac{1}{n}\ (s_1 +\dots + s_N)+ \frac{1}{n}(\alpha - \epsilon)(n-N)\\\implies \sigma_n &\geq \frac{1}{n}\ (s_1 + \dots + s_N)- \frac{N}{n}(\alpha - \epsilon)+\alpha - \epsilon\end{align}
Now, $\frac1n(s_1 + \dots+ s_N)-\frac{N}{n}(\alpha-\epsilon)<\epsilon_1$ as $n \rightarrow \infty$ arbitrarily small, so $$\forall \ \epsilon_1>0 \ \exists N_1: \forall n>N_1: \sigma_n>\alpha-\epsilon-\epsilon_1$$ Now I really don't know how to infer from it that $\liminf_{n→∞}  s_n \leq \liminf_{n→∞} σ_n$.
So any help will be much appreciated!

Note: this is a response to the first version whose details were a little hard to read. 
Fix any $N$. For any $n>N$ we have 
$$\sigma_n =\frac 1n (s_1+\dots+s_n) \ge \frac 1 n ( s_1+\dots+ s_N + (n-N) \min\{s_{N+1},\dots,s_n\}) \ge \frac{s_1+\dots+s_N}{n} + \frac{n-N}{n} \inf \{s_{N+1},s_{N+2},\dots\}.$$
Therefore $$\liminf_{n\to\infty} \sigma_n \ge 0 + \inf \{s_{N+1},s_{N+2},\dots\}.$$
This is true for every $N$. The RHS increases to $\liminf_{n\to\infty} s_n$, as $n\to\infty$, and the result follows. 
