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.

  • 1
    $\begingroup$ $1 \leq 3$, $1 \times 2 \leq 3 \times 2$, $1 \times 2 \times 3 = 3 \times 2 \times 1$ - how is your example a counterexample? $\endgroup$ – Soarer Jul 22 '11 at 7:47
  • 5
    $\begingroup$ @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. $\endgroup$ – Martin Sleziak Jul 22 '11 at 7:57
  • 1
    $\begingroup$ @Martin. +1. Thanks Martin. I appreciate it. $\endgroup$ – Henry B. Jul 22 '11 at 8:01
  • 4
    $\begingroup$ @Henry Prof. Steele collects typos and errors at www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/… You should drop him a line! $\endgroup$ – user940 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 :-(

| cite | improve this answer | |
  • $\begingroup$ It seems convincing enough to me. $\endgroup$ – Henry B. Jul 22 '11 at 8:09
  • 1
    $\begingroup$ 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. $\endgroup$ – LostInMath Jul 22 '11 at 12:27
  • $\begingroup$ Martin, I hope you don't mind my minor corrections. $\endgroup$ – Rasmus Jul 25 '11 at 19:35
  • $\begingroup$ Thanks @Rasmus! I should read my answers more carefully after typing them ;-) $\endgroup$ – Martin Sleziak Jul 25 '11 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.