Prove that $\frac{a_1}{a_9} + \frac{a_2}{a_8}+...\frac{a_9}{a_1} \le 1 + \frac{4(p^2+q^2)}{pq}$

Question

Let $a_1, a_2, ..., a_9$ be a sequence of numbers satisfying $0<p \le a_i \le q$ for each $i = 1,2,...,9.$ Prove that $$\frac{a_1}{a_9} + \frac{a_2}{a_8}+...\frac{a_9}{a_1} \le 1 + \frac{4(p^2+q^2)}{pq} .$$

I'm having quite a hard time with this one. Writing out everything on the LHS of the inequality given above, we have $$\frac{a_1}{a_9} + \frac{a_2}{a_8}+ \frac{a_3}{a_7} + \frac{a_4}{a_6}+ \frac{a_5}{a_5}+ \frac{a_6}{a_4}+ \frac{a_7}{a_3}+ \frac{a_8}{a_2}+ \frac{a_9}{a_1},$$ which is equivalent to $$\frac{a_1}{a_9} + \frac{a_2}{a_8}+ \frac{a_3}{a_7} + \frac{a_4}{a_6}+ 1+ \frac{a_6}{a_4}+ \frac{a_7}{a_3}+ \frac{a_8}{a_2}+ \frac{a_9}{a_1}.$$ Supposing the sequence increases monotonically, to the left of 1 all terms are closer to $\frac{p}{q}$ and to the right of 1 all terms are closer in value to $\frac{q}{p}$. Therefore, the long list of fractions above is approximately equivalent to $1 + 4(\frac{p}{q} + \frac{q}{p}) = 1+\frac{4(p^2+q^2)}{pq}$. While I am underestimating on the left of $1$, I am overestimating on the right of $1$. But since clearly $\frac{q}{p} > 1$, the overall expression $1+\frac{4(p^2+q^2)}{pq}$ will be greater than the actual total sum.

If the sequence were decreasing monotonically, we could still use a similar approach and end up with the same answer. However, I'm worried about my solution because I can't seem to justify my assumption that the sequence is either monotonically increasing/decreasing — please help!

Answer 1

Let $n$ be an arbitrary positive integer. Define $$f\left(a_1,a_2,\ldots,a_n\right):=\sum_{i=1}^n\,\frac{a_i}{a_{n+1-i}}$$ for $a_1,a_2,\ldots,a_n\in [p,q] \subseteq (0,\infty)$. Observe that $\frac{p}{q}\leq \frac{a_i}{a_{j}}\leq \frac{q}{p}$ for every $i,j\in\{1,2,\ldots,n\}$. Therefore, for $i=1,2,\ldots,\left\lfloor\frac{n}{2}\right\rfloor$, $$\frac{q}{p}\left(\frac{a_i}{a_{n+1-i}}-\frac{p}{q}\right)\left(\frac{a_{n+1-i}}{a_i}-\frac{p}{q}\right)\geq 0\text{ or }\frac{a_i}{a_{n+1-i}}+\frac{a_{n+1-i}}{a_i}\leq \frac{p}{q}+\frac{q}{p}\,.$$ Hence, $$f\left(a_1,a_2,\ldots,a_n\right)\leq \epsilon_n+\sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\,\left(\frac{p}{q}+\frac{q}{p}\right)=\epsilon_n+\left\lfloor\frac{n}{2}\right\rfloor\left(\frac{p}{q}+\frac{q}{p}\right)\,,$$ where $\epsilon_n:=0$ for even $n$, and $\epsilon_n:=1$ for odd $n$. The maximum value is achieved if and only if $$\left\{a_i,a_{n+1-i}\right\}=\{p,q\}$$ for every $i=1,2,\ldots,\left\lfloor\frac{n}{2}\right\rfloor$.

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Interesting, Related Problem

Let $n$ be a positive integer, and $\sigma$ a permutation on the set $\{1,2,\ldots,n\}$. Define $$f_\sigma\left(a_1,a_2,\ldots,a_n\right):=\sum_{i=1}^n\,\frac{a_i}{a_{\sigma(i)}}\,,$$ for $a_1,a_2,\ldots,a_n\in [p,q]\subseteq (0,\infty)$. Show that the maximum possible value of $f$ is $$\sum_{j=1}^k\,\Biggl(\epsilon_{t_k}+\left\lfloor\frac{t_k}{2}\right\rfloor\left(\frac{p}{q}+\frac{q}{p}\right)\Biggr)=n+\left(\sum_{j=1}^k\,\left\lfloor\frac{t_k}{2}\right\rfloor\right)\,\frac{(p-q)^2}{pq}\,,$$ where $\epsilon_{r}:=0$ for each even integer $r$ and $\epsilon_r:=1$ for each odd integer $r$. Here, $t_1,t_2,\ldots,t_k$ are the lengths of the cycles in the decomposition of $\sigma$ into a product of disjoint cyclic permutations. The OP's problem is a special case of my generalized version where $n=9$ and $$\sigma=(1\;\;9)(2\;\;8)(3\;\;7)(4\;\;6)(5)\,,$$ so that $k=5$, $t_1=t_2=t_3=t_4=2$, and $t_5=1$

Answer 2

The hint.

Let $f(a_1,a_2,...,a_9)=\sum\limits_{k=1}^9\frac{a_k}{a_{10-k}}$, where $a_0=a_9$.

Thus, $$\frac{\partial^2f}{\partial a_k^2}>0,$$ which says that $f$ is a convex function of $a_k$ for all $k$.

Thus, $$\max_{a_k\in[p,q]}f=\max_{a_k\in\{p,q\}}f$$