Showing $\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n b_i \right)^\frac1{n}\le\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n a_i \right)^\frac1{n}$?

If $a_1\ge a_2 \ge a_3 \ldots$ and if $b_1,b_2,b_3\ldots$ is any rearrangement of the sequence $a_1,a_2,a_3\ldots$ then for each $N=1,2,3\ldots$ one has

$$\sum^N_{n=1}\left(\prod_{i=1}^n b_i \right)^{\frac{1}{n}}\le \sum^N_{n=1}\left(\prod_{i=1}^n a_i \right)^{\frac{1}{n}}$$

This comes from page 177 of "The Cauchy-Schwarz Master Class".

The solution in the back argues that, by hypothesis, $b_1\le a_1,b_2\le a_2,b_3\le a_3\dots$ Therefore, it follows that $(b_1b_2\cdots b_n)^{1/n}\le (a_1a_2\cdots a_n)^{1/n}$.

It seems to me that for $N=3$, with a sequence $a_1=3$,$a_2=2$ and $a_3=1$, and it's rearrangement $b_1=1$,$b_2=2$ and $b_3=3$, this is not the case.

Am I missing something obvious?

In order to provide the context, here is the relevant portion from the book (Steele J.M. The Cauchy-Schwarz master class, CUP 2004, p.273):

Solution for Exercise 11.7. This observation is painfully obvious, but it seems necessary for completeness. The hypothesis gives us the bounds $b_1 \le a_1, b_2 \le a_2, \dots , b_N \le a_N$; thus, for all $1 \le n \le N$ we have $(b_1b_2\dots b_n)^{1/n} \le (a_1a_2\dots a_n)^{1/n}$, which is more than we need. There are questions on infinite rearrangements which are subtle, but this is not one of them.

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$1 \leq 3$, $1 \times 2 \leq 3 \times 2$, $1 \times 2 \times 3 = 3 \times 2 \times 1$ - how is your example a counterexample? – Soarer Jul 22 '11 at 7:47
@Soarer: It is not a counterexample to the inequality, but to the claim, which is made in the solution given in the book. Henry: I've edited your question - I've copied the relevant part from the book. I hope you don't mind. – Martin Sleziak Jul 22 '11 at 7:57
@Martin. +1. Thanks Martin. I appreciate it. – Henry B. Jul 22 '11 at 8:01
@Henry Prof. Steele collects typos and errors at www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/… You should drop him a line! – Byron Schmuland Jul 22 '11 at 12:59

I think you are correct with your observation.

Maybe the author wanted to say that for each $n$ such that $1\le n\le N$ we have

$$a_1a_2\dots a_n \ge b_1b_2 \dots b_n,$$

which follows from the fact that if we reorder $b_1,b_2,\ldots,b_n$ from the largest element $c_1\ge c_2\ge\ldots\ge c_n$, then $c_1\le a_1,c_2\le a_2,\ldots,c_n\le a_n$ and $b_1b_2\dots b_n = c_1c_2\dots c_n$.

Although this observation seems to be easy, I have trouble writing a simple and clear proof of it :-(

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It seems convincing enough to me. – Henry B. Jul 22 '11 at 8:09
While the claim $c_i \leq a_i,i=1\ldots n$ is intuitively very clear, one way to prove it rigorously is to note that $(c_i)_{i=1}^n$ is a subsequence of $(a_i)_{i=1}^N$ (i.e. there are indices $1\leq k_1 < k_2 < \cdots < k_n \leq N$ s.t. $c_i=a_{k_i},i=1,\ldots,n$. This implies that $c_i=a_{k_i}\leq a_i$, because $k_i \geq i$. One way to find these indices $k_i$ is to define inductively $k_i=\min\{k: k_{i-1}<k\leq N\text{ and }c_i=a_k\}$. This definition makes sense (i.e. the set after $\min$ is nonempty) because both the sequences $(a_i)_{i=1}^N$ and $(c_i)_{i=1}^n$ are nonincreasing. – LostInMath Jul 22 '11 at 12:27
Martin, I hope you don't mind my minor corrections. – Rasmus Jul 25 '11 at 19:35
Thanks @Rasmus! I should read my answers more carefully after typing them ;-) – Martin Sleziak Jul 25 '11 at 19:41