# Upper bound of $x_1x_2x_3+x_2x_3x_4+\dots+x_{n}x_1x_2$

Let $n\geq 3$ be a positive integer and let $x_i$'s be non-negative real numbers with $x_1+x_2+\dots+x_n=1$. What is the maximum of $x_1x_2x_3+x_2x_3x_4+\dots+x_{n}x_1x_2$?

If the sum were symmetric consisting of all terms of the form $x_ix_jx_k$ for $i<j<k$, then equality would hold at $x_1=\dots=x_n=1/n$. For a cyclic sum, however, this might not be true. As an example, for the sum $x_1x_2+x_2x_3+\dots+x_nx_1$ the maximum is $1/4$, attained when $x_1=x_2=1/2$ and the rest of $x_i$ are $0$.

• If $n=3$, the summation is $x_1x_2x_3 + x_2x_3x_1$? – MCT Jul 20 '16 at 20:34
• If $n=3$ it is $x_1x_2x_3+x_2x_3x_1+x_3x_1x_2=3x_1x_2x_3$, so the max is attained when $x_1=x_2=x_3=1/3$, and is $1/9$. – pi66 Jul 20 '16 at 20:35
• Sorry, you raised a good point. The sum should go to $x_nx_1x_2$. – pi66 Jul 20 '16 at 20:36
• If $n \ge 5$, $x_1 = x_2 = x_3 = 1/3$, and all other $x_i$ are zero, your sum is $1/27$. This beats the $1/n^2$ you'd get if $x_1 = \cdots = x_n = 1/n$ for $n \ge 6$. – Michael Lugo Jul 20 '16 at 20:45
• @Michael Lugo for $n=5$ a maximum is $\frac{1}{25}$. – Michael Rozenberg Jul 20 '16 at 21:46

1. Let $n=3$. Hence, by AM-GM we get $x_1x_2x_3+x_2x_3x_2+x_3x_1x_2\leq3\left(\frac{x_1+x_2+x_3}{3}\right)^3=\frac{1}{9}$.

The equality occurs for $x_1=x_2=x_3=\frac{1}{3}$. Id est, the answer is $\frac{1}{9}$.

1. Let $n=4$. Hence, by AM-GM $x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2=$

$=x_1x_2(x_3+x_4)+x_3x_4(x_1+x_2)\leq\left(\frac{x_1+x_2}{2}\right)^2(x_3+x_4)+\left(\frac{x_3+x_4}{2}\right)^2(x_1+x_2)=$ $=\frac{1}{4}(x_1+x_2)(x_3+x_4)\leq\frac{1}{4}\left(\frac{x_1+x_2+x_3+x_4}{2}\right)^2=\frac{1}{16}$.

The equality occurs for $x_1=x_2=x_3=x_4=\frac{1}{4}$. Is est, the answer is $\frac{1}{16}$.

1. Let $n=5$ and $x_1=\min\{x_1,x_2,x_3,x_4,x_5\}$.

Hence, by AM-GM again $x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+x_4x_5x_1+x_5x_1x_2=$

$=x_1(x_2+x_4)(x_3+x_5)+x_3x_4(x_2+x_5-x_1)\leq$

$\leq x_1\left(\frac{x_2+x_3+x_4+x_5}{2}\right)^2+\left(\frac{x_2+x_3+x_4+x_5-x_1}{3}\right)^3=x_1\left(\frac{1-x_1}{2}\right)^2+\left(\frac{1-2x_1}{3}\right)^3\leq\frac{1}{25}$

because the last inequality it's $(5x_1-1)^2(5x_1+8)\geq0$.

The equality occurs for $x_1=x_2=x_3=x_4=x_5=\frac{1}{5}$. Id est, the answer is $\frac{1}{25}$.

1. Let $n\geq6$.

a) $n=3k$, where $k\in\mathbb N$ and $x_{n+1}=x_1$, $x_{n+2}=x_2$.

Hence, by AM-GM $\sum\limits_{i=1}^nx_ix_{i+1}x_{i+2}\leq(x_1+x_4+...)(x_2+x_5+...)(x_3+x_6+...)\leq$

$\leq\left(\frac{x_1+x_2+...+x_n}{3}\right)^3=\frac{1}{27}$.

b) $n=3k+1$, where $k\in\mathbb N$ and $x_1=\min\{x_1,x_2,...x_n\}$.

Hence, by AM-GM $\sum\limits_{i=1}^nx_ix_{i+1}x_{i+2}\leq(x_1+x_4+...)(x_2+x_5+...)(x_3+x_6+...)\leq$

$\leq\left(\frac{x_1+x_2+...+x_n}{3}\right)^3=\frac{1}{27}$.

c) $n=3k+2$, where $k\in\mathbb N$ and $x_{n-2}=\max\{x_1,x_2,...x_n\}$.

Hence, by AM-GM $\sum\limits_{i=1}^nx_ix_{i+1}x_{i+2}\leq(x_1+x_4+...)(x_2+x_5+...)(x_3+x_6+...)\leq$

$\leq\left(\frac{x_1+x_2+...+x_n}{3}\right)^3=\frac{1}{27}$.

The equality occurs for $x_1=x_2=x_3=\frac{1}{3}$ and $x_4=x_5=...=x_n=0$. Id est, the answer is $\frac{1}{27}$.

Done!