Average Inequality I came across the following inequality:

In an ordered field, suppose $a_1 \leq a_2 \leq \dots$ Let $b_n = (a_1+ \dots + a_n)/n$. Show that $b_1 \leq b_2 \leq b_3 \leq \dots$

This is equivalent to showing that $$a_1 \leq \frac{a_1+a_2}{2} \leq \frac{a_1+a_2+a_3}{3} \leq \dots$$
Would the easiest way to proceed be by induction on $n$? 
 A: Don't think you really need to use induction. Using the assumption to prove each successive equality independently may be easier.
A: Don't know if it is "the easiest", but it's certainly a pretty straightforward way of doing it. It's a little simpler to prove
$$(n+1)(a_1+\cdots+a_n) \leq n(a_1+\cdots+a_{n+1})$$
which is equivalent.
The case $n=1$ is easy (as usual): $2a_1 = a_1+a_1\leq a_1+a_2 = 1(a_1+a_2)$, so you are set.
Assuming the case $n=k$ holds, you want to establish the case $n=k+1$; you assume
$$(k+1)(a_1+\cdots+a_k) \leq k(a_1+\cdots + a_{k+1})$$
and you want to show that
$$(k+2)(a_1+\cdots+a_{k+1})\leq (k+1)(a_1+\cdots+a_{k+2}).$$
You have:
$$\begin{align*}
(k+2)(a_1+\cdots+a_{k+1}) &= (k+1)(a_1+\cdots+a_k) + (a_1+\cdots+a_k) + (k+2)a_{k+1}\\
&\leq k(a_1+\cdots+a_{k+1}) + (a_1+\cdots+a_k) + (k+2)a_{k+1}\\
&= (k+1)(a_1+\cdots+a_k) + ka_{k+1} + (k+2)a_{k+1}\\
&= (k+1)(a_1+\cdots +a_k) + (k+1)a_{k+1} + (k+1)a_{k+1}\\
&= (k+1)(a_1+\cdots+a_{k+1}) + (k+1)a_{k+1}\\
&\leq (k+1)(a_1+\cdots+a_{k+1}) + (k+1)(a_{k+2})\\
&= (k+1)(a_1+\cdots+a_{k+2}),
\end{align*}$$
Which proves the induction. 
A: We have $a_i \leq a_j$ whenever $i \leq j$.
Use induction to show that $a_1 + a_2 + a_3 + \cdots + a_n \leq n a_{n+1}$
Now add $n \times (a_1 + a_2 + a_3 + \cdots + a_n)$ to both sides to get
$$n \times (a_1 + a_2 + a_3 + \cdots + a_n) +  (a_1 + a_2 + a_3 + \cdots + a_n) \leq n \times (a_1 + a_2 + a_3 + \cdots + a_n) + n a_{n+1}$$ and now rearrange to get
$$\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} \leq \frac{a_1 + a_2 + a_3 + \cdots + a_n + a_{n+1}}{n+1}$$
A: I think so.  Your base case is the first inequality-can you prove that?  Then, since $\le$ is transitive, $a_{n+1} \ge \frac {1}{n} \sum_{i=1}^n a_n$
