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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 $n$?

  • 1
    Induction should work quite nicely.2011-06-22

4 Answers 4

0

I think so. Your base case is the first inequality-can you prove that? Then, since $\le$ is transitive, $a_{n+1} \ge \frac {1}{n} \sum_{i=1}^n a_n$

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Don't think you really need to use induction. Using the assumption to prove each successive equality independently may be easier.

  • 0
    @Didier, you may be right, in which case, Steve's hint was simply too elliptical for me.2011-06-29
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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.

  • 0
    Love those hit-and-run...2011-06-29
1

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

$\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} \leq \frac{a_1 + a_2 + a_3 + \cdots + a_n + a_{n+1}}{n+1}$