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Assume $0 < p_1 \le p_2 \le \dots \le p_{2k}$. I am looking for a (preferably tight) upper bound for the following expression:

$$ \frac {(p_1+p_{2k})(p_2+p_{2k-1})\dots(p_{k-1}+p_{k+2})(p_k+p_{k+1})} {(p_1+p_{k+1})(p_2+p_{k+2})\dots(p_{k-1}+p_{2k-1})(p_k+p_{2k})}. $$

Note: I have already asked about a special case of this question here. However, since the general case may require a different approach to solve (or maybe, there is no answer for it), I prefer to ask a new question.

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1 Answer 1

up vote 2 down vote accepted

The expression depends on $p_1$ only through

$$\frac{p_1+p_{2k}}{p_1+p_{k+1}}\;.$$

Since $0\lt p_1\le p_{k+1}\le p_{2k}$, this is maximal for $p_1=0$. The same argument applies for all $p_i$ up to and including $p_n$ with $n:=\lceil k/2\rceil$. That leaves us with

$$ \frac {p_{2k}\dots p_{2k-n+1}(p_{n+1}+p_{2k-n})\dots(p_k+p_{k+1})} {p_{k+1}\dots p_{k+n}(p_{n+1}+p_{k+n+1})\dots(p_k+p_{2k})}\;. $$

Now the factor involving $p_{2k}$ is

$$\frac{p_{2k}}{p_k+p_{2k}}\;.$$

Since $0\lt p_k\le p_{2k}$, this is less than $1$ and we can make it arbitrarily close to $1$ by making $p_{2k}$ arbitrarily large. Thus we can drop this factor for the purpose of determining the supremum of the expression. The same argument applies to all $p_i$ down to and including $p_{2k-n+1}$. Since $2k-n+1\le k+n+1$, this gets rid of all sums in the denominator and leaves us with

$$ \frac {(p_{n+1}+p_{2k-n})\dots(p_k+p_{k+1})} {p_{k+1}\dots p_{2k-n}}\;. $$

This expression is maximal if $p_{n+1}$ through $p_k$ are as large as possible and $p_{k+1}$ through $p_{2k-n}$ are as small as possible, and thus, taking into account the constraints, if they are all equal. Thus the desired supremum of the expression is $2^{k-n}=2^{\lfloor k/2\rfloor}$.

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