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Consider positive numbers $ x_0,\ldots, x_{n-1}$, and extend the sequence cyclicly by setting $ x_i=x_{i\,{\rm mod}\,n}$. Now, for $1\le k\le n$ we define $ f_k(t_0,t_1,\ldots, t_{k-1})=\prod_{j=0}^{k-1}t_j$ and we set $$F_{k,n}(x_0,x_1,\ldots,x_{n-1})=\sum_{j=0}^{n-1}f_k(x_j,x_{j+1},\ldots,x_{j+k-1})$$ I am considering $$M_{n,k}=\sup_{x_0+\cdots+x_{n-1}=1}F_{k,n}(x_0,x_1,\ldots,x_{n-1})$$ Testing sequences of the form $x_0=\cdots=x_{k-1}=1/k$ and $x_0=\cdots=x_{n-1}=1/n$ Shows that $$M_{n,k}\ge 1/\min( n^{k-1},k^k).\tag{$*$}$$ It is not hard, to show that we have $M_{n,k}=k^{-k}$ when $1\le k\le n/2$, so we have equality in $(*)$ in this case. Also, it is easy to prove equality in $(*)$ when $k=n,n-1$.

The remaining cases seem intractable to me.

My question is: Does equality in $(*)$ hold when $\frac{n}{2}< k\le n-2$ ?

Felix Klein
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1 Answers1

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There is the following example: $n=5$ and $k=3$.

Let $a$, $b$, $c$, $d$ and $e$ be positive numbers such that $a+b+c+d+e=1$. Prove that: $$abc+bcd+cde+dea+eab\leq\frac{1}{25}$$

Indeed, let $e=\min\{a,b,c,d,e\}$. Hence, $e\leq\frac{1}{5}$ and by AM-GM we obtain: $$abc+bcd+cde+dea+eab=e(a+c)(b+d)-bc(a+d-e)\leq$$ $$\leq e\left(\frac{a+c+b+d}{2}\right)^2+\left(\frac{b+c+a+d-e}{3}\right)^3=$$ $$=\frac{e(1-e)^2}{4}+\frac{(1-2e)^3}{27}.$$ Thus, it's enough to prove that $$\frac{e(1-e)^2}{4}+\frac{(1-2e)^3}{27}\leq\frac{1}{25}$$ or $$(5e-1)^2(5e+8)\geq0$$ and we are done!

Also for $n=6$ and $k=4$ the equality is true, but my proof of this statement is very ugly.

See here:How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$?

For $n>6$ I have no any example, but it seems that this inequality is true.

Michael Rozenberg
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