If $(a_{n})_{n}$ is a bounded sequence, show that $\liminf_{n\to \infty}a_{n}\leq \liminf_{n\to \infty}\frac{a_{1}+a_{2}+\cdots +a_{n}}{n}$. I am very lost at this proof. I tried to break it a part but it made it more confusing. I am not sure what algebra or theorems I must use.

We have $\liminf_{n}a_n = \sup_{m \geq 1}\inf_{n\geq m}a_n$
Suppose that $\liminf _n \dfrac{a_1+a_2+\cdots +a_n}{n} = a$. Then for all $m$
$$\liminf _n \dfrac{a_m+a_{m+1}+\cdots +a_n}{n} = a.$$ 
Notice that $\dfrac{a_m+a_{m+1}+\cdots +a_n}{n}\geq (n-m+1)\frac{\min(a_m,a_{m+1},\ldots, a_n)}{n}$, taking $\liminf$ we get 
$$a\geq \inf_{n\geq m}a_n$$
Now, taking $\sup$ over all $m$ we obtain 
$$a\geq \sup_{m \geq 1}\inf_{n\geq m}a_m=\liminf_{n}a_n$$
