calculus - Prove $b_n$ is increasing and bounded above given $a_n$ is increasing and bounded above.

If $\{a_n\}$, $n \ge 1$, is increasing and bounded above, then so is $\{b_n\}$, $n \ge 1$, where $b_n = \frac{a_1+a_2+...+a_n}{n}$.

So far, I understand that for a sequence to be increasing, $b_{n+1} \ge b_n$, but every time I try to algebraically manipulate this statement, I can't seem to prove that to be the case.

An attempt:

$$\frac{a_1+a_2+...+a_n+a_{n+1}}{n+1} \ge \frac{a_1+a_2+...+a_n}{n}$$
$$\frac{n}{a_1+a_2+...+a_n} * \frac{a_1+a_2+...+a_n+a_{n+1}}{n+1} \ge 1$$
$$\frac{n}{n+1} + \frac{n(a_{n+1})}{(n+1)(a_1+a_2+...+a_n)} \ge \frac{n+1}{n+1}$$
$$\frac{n(a_{n+1})}{(n+1)(a_1+a_2+...+a_n)} \ge \frac{1}{n+1}$$
$$a_{n+1} \ge \frac{a_1+a_2+...+a_n}{n}$$

As for proving an upper bound exists, I am pretty stumped on how I could prove that for this situation.