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 $C?$



Find the best constant to this bounded inequality
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 $C\geq 7/12$.


