# 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 $C?$

• What rules out the case of $x_1=2$ (or $C+1$) and the others set to zero or something really small and positive? Gerhard "Not Sure It's Been Proven" Paseman, 2016.07.09. – Gerhard Paseman Jul 10 '16 at 1:07
• @GerhardPaseman Don't the conditions imply $x_1^2 \leq 1$? – Todd Trimble Jul 10 '16 at 1:26
• If $i=j$ is allowed on the constraints, then it would rule out my suggested example. Although the phrasing is explicit, I would add your implication as a consequence to the conditions to make the problem robust against mistakes like mine. Gerhard "Back For A Third Reading" Paseman, 2016.07.09. – Gerhard Paseman Jul 10 '16 at 1:32
• May be start with simple $(\sum x_k)^2=\sum x_k x_m$ -? – Sergei Jul 13 '16 at 12:55

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 $C\geq 7/12$.
@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 $x_1+x_2+\dots+x_{2m+1}=\frac{5}{3}-\frac{2}{3\cdot 4^m}$.
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.
• @T.Amdeberhan - you are right with $\sqrt{\frac{5n}{3}}$, I mean arithmetic mean by mistake, thank you. – Sergei Jul 15 '16 at 4:54