Find the best constant to this bounded inequality Let $n$ be postive integer number, and $x_{i}\ge 0$, such 
$$x_{i}x_{j}\le 4^{-|i-j|},1\le i,j\le n$$
then I have prove 
$$x_{1}+x_{2}+\cdots+x_{n}<\dfrac{5}{3}$$
Edit Add Proof:since $x^2_{i}\le 1,0\le x_{i}\le 1$,Let $S_{j}=\sum_{i=1}^{j}x_{i},S=\sum_{i=1}^{n}x_{i}$,then we have
$$0=S_{0}\le S_{1}\le S_{2}\le\cdots\le S$$so there exist $k$,such
$S_{k}\le\dfrac{S}{2}\le S_{k+1}$,
if we let $$T_{k}=S-S_{k},T_{k+1}=S-S_{k+1}$$
then we have
$$|S_{k}-T_{k}|+|S_{k+1}-T_{k+1}|=|2S_{k}-S|+|2S_{k+1}-S|=2x_{k+1}\le 2$$
then for $l\in\{k,k+1\}$,we have
$$|S_{l}-T_{l}|\le 1\tag{1}$$
and we have
$$S_{l}T_{l}=\sum_{i=1}^{l}\sum_{j=l+1}^{n}x_{i}x_{j}\le\sum_{i=1}^{l}\sum_{j=l+1}^{n}4^{-|i-j|}\le\sum_{i=1}^{l}\dfrac{1}{4^{l-i}}\sum_{j=l+1}^{n}\dfrac{1}{4^{j-l}}<\dfrac{4}{3}\cdot\dfrac{1}{3}\tag{2}$$
use $(1)$ and $(2)$
we have
$$x_{1}+x_{2}+\cdots+x_{n}=S_{l}+T_{l}=\sqrt{(S_{l}-T_{l})^2+4S_{l}T_{l}}\le\dfrac{5}{3}$$
Question :
Let $n$ be postive integer number, and $x_{i}\ge 0$, such 
$$x_{i}x_{j}x_{k}\le 4^{-|i-j-k|},1\le i,j,k\le n$$
then I have prove 
$$x_{1}+x_{2}+\cdots+x_{n}<C$$
find the best constant  $C?$
 A: These are some thoughts concerning the Question. We have that
\begin{align*}
&S_n(3)=\sum_{1\leq i\leq n}x_i^3< \sum_{i\geq 1}4^{-i}=1/3,\\
&S_n(1,2)=\sum_{1\leq i<j\leq n}x_i x_j^2< \sum_{1\leq i<j}4^{-(2j-i)}=1/45,\\
&S_n(2,1)=\sum_{1\leq i<j\leq n}x_i^2 x_j< \sum_{1\leq i<j}4^{-j}=1/9,\\
&S_n(1,1,1)=\sum_{1\leq i<j<k\leq n}x_i x_j x_k< \sum_{1\leq i<j<k}4^{-(k+j-i)}=1/135.
\end{align*}
Hence
$$(x_1+x_2+\dots+x_n)^3=3(2S_n(1,1,1)+S_n(2,1)+S_n(1,2))+S_n(3)<\frac{7}{9}$$
which means that $C\leq \sqrt[3]{7/9}$.
On the other hand, note that $x_1=1/4$ and $x_i=1/4^{i-1}$ for $2\leq i\leq n$ satisfy the required conditions and 
$$x_1+x_2+\dots+x_n=\frac{7}{12}-\frac{1}{3\cdot 4^{n-1}}.$$
Therefore $C\geq 7/12$.
A: @T.Amdeberhan - I think that the constant $\frac{5}{3}$ is optimal.
Take $x_1=1/4^{m}$, $x_2=1/4^{m-1}$,$\dots$ ,$x_m=1/4$, $x_{m+1}=1$, $x_{m+2}=1/4$, $\dots$, $x_{2m}=1/4^{m-1}$, $x_{2m+1}=1/4^m$, then $x_ix_j\leq 1/4^{|i-j|}$ and $x_1+x_2+\dots+x_{2m+1}=\frac{5}{3}-\frac{2}{3\cdot 4^m}$.
A: Start with $(\sum x_k)^2=\sum x_k^2 + 2\sum_{i>j}x_ix_j .$
Under given conditions and summing up progressions give
$$(\sum x_k)^2 \le \frac{5n}{3} -\frac{2}{9}(1-(\frac{1}{4})^n).
$$
So
$$\sum x_k \le \frac{5}{3n}.$$
This seems to be true without condition $x_k>0$ with modulo sigh in lhs. May we do the same with products of 3 x-s and so on?
Another consideration which may be useful. There is a known inequality for values $\sum x_k, \sum_{i>j}x_ix_j$, but with the wrong sigh for our problem (attributed to Newton). May be there is also an analogue with opposite sigh, but I do not know it.
