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.

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 :-(