Let $n=3k+r$ where $0\le r<3$ and note that the expression is linear in $\{x_1,x_4,\ldots,x_{3k-2}\}$, $\{x_2,x_5,\ldots,x_{3k-1}\}$, and $\{x_3,x_6,\ldots,x_{3k}\}$. Thus by convexity we have that at most one number from each of those sets is nonzero. If $k>1$, then we will have $5\ge r+3\ge k$ nonzero numbers spaced around the circle. Label them $a_1,a_2,\ldots,a_k$ so that $a_1$ and $a_k$ are not adjacent, which is possible as $k\le r+3<n$. Now, note that it is optimal to 'push' all the nonzero numbers together, so the expression is at most $a_1a_2a_3+a_2a_3a_4+a_3a_4a_5$ where the nonzero numbers are among $a_1,a_2,a_3,a_4,a_5$. Now, as this expression is linear in $\{a_1,a_4\}$ and $\{a_2,a_5\}$, we have by convexity again that two of these are 0, so pushing the remaining three nonzero numbers together again, we have by AM-GM that the maximum is $\frac1{27}$. When $n=4$, we can note that the expression is symmetric and that replacing two of the numbers by their average increases the expression unless they are equal, so the maximum occurs at $x_1=x_2=x_3=x_4=\frac14$, leading to a maximum of $\frac1{16}$. The case of $n=5$ is much more annoying, and it is likely that Lagrange multipliers gives the cleanest solution. A sketch: after getting rid of edge cases and taking partials, we want $x_1(x_5+x_2)=x_2(x_1+x_3)=x_3(x_2+x_4)=x_4(x_3+x_5)=x_5(x_4+x_1)$. Treating this as a system of equations in $x_1x_2,x_2x_3,x_3x_4,x_4x_5,x_5x_1$, we have that they are all equal and thus $x_1=x_2=x_3=x_4=x_5$, leading to a maximum of $\frac1{25}$.

ABCDE wrote: Let $n=3k+r$ where $0\le r<3$ and note that the expression is linear in $\{x_1,x_4,\ldots,x_{3k-2}\}$, $\{x_2,x_5,\ldots,x_{3k-1}\}$, and $\{x_3,x_6,\ldots,x_{3k}\}$. Thus by convexity we have that at most one number from each of those sets is nonzero. I am sorry, but why is this true? How would you deal with the condition $x_1 + \cdots + x_n = 1$? Thanks!

Let $x_1\ge x_2\ge... \ge x_n$ So $x_1x_2x_3 + x_2x_3x_4 + \cdots + x_nx_1x_2 \le x_1x_2x_3+x_1x_2x_4+...+x_nx_1x_2=x_1x_2(1-x_1-x_2)\le \frac{(x_1+x_2+1-x_1-x_2)^3}{27}=\frac{1}{27}$ Who can check it? İt seems wrong?

@above: Since the inequality is cyclic, you cannot fix the order of the elements. You can only do so if it is a symmetric inequality. @MathPanda1: If we say fix all the variables except those of $\{x_1,x_4,\ldots,x_{3k-2} \}$, then the inequality then becomes of the form $x_1 + x_4 + \cdots + x_{3k-2} = r$ for some non-negative real $r$ and we have to minimize the expression $x_1c_1 + x_4c_2 + \cdots +x_{3k-2}c_k$ where the $c_i$'s are constants. Then it is clear that the minimum value is $rc_j$ where $c_j$ is the smallest among the $c_i$'s. Hence we can assume that at most one element of this set is non-zero.

fattypiggy123 wrote: Let $n \geq 4$ be a natural and let $x_1,\ldots,x_n$ be non-negative reals such that $x_1 + \cdots + x_n = 1$. Determine the maximum value of $x_1x_2x_3 + x_2x_3x_4 + \cdots + x_nx_1x_2$. See here: http://math.stackexchange.com/questions/2188952