Find the maximum of the value $F=x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+\cdots+x_{n-2}x_{n-1}x_{n}+x_{n-1}x_{n}x_{1}$ 
Let $x_{i}\ge 0$ and such $$x_{1}+x_{2}+\cdots+x_{n}=1$$
  Find the maximum of the value
  $$F=x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+\cdots+x_{n-2}x_{n-1}x_{n}+x_{n-1}x_{n}x_{1}$$

case $1$. when $n=3$,then
$$F=2x_{1}x_{2}x_{3}\le2\dfrac{(x_{1}+x_{2}+x_{3})^3}{27}=\dfrac{2}{27}$$
but for $n\ge 4$,I can't solve it 
 A: I think you meant that $F=x_1x_2x_3+x_2x_3x_4+...+x_nx_1x_2$, 
but I am ready to think also about your $F$.
For $n=4$ we'll prove that $abc+abd+acd+bcd\leq4$, 
where $a$, $b$, $c$ and $d$ are non-negatives such that $a+b+c+d=4$,
Indeed, by AM-GM
$$abc+abd+acd+bcd=ab(c+d)+cd(a+b)\leq\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)=$$
$$=\frac{(a+b)(c+d)(a+b+c+d)}{4}=(a+b)(c+d)\leq\left(\frac{a+b+c+d}{2}\right)^2=4$$
Thus, $$abc+abd+acd+bcd\leq4\left(\frac{a+b+c+d}{4}\right)^3,$$
which says that $$x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2\leq4\left(\frac{x_1+x_2+x_3+x_4}{4}\right)^3=\frac{1}{16}.$$
The equality occurs for $x_1=x_2=x_3=x_4=\frac{1}{4}$, which says that the answer is $\frac{1}{16}$.
For $n=5$ the answer is $\frac{1}{25}$.
Indeed, let $x_1=a$, $x_2=b$, $x_3=c$, $x_4=d$,$x_5=e$ and $a=b=c=d=e=\frac{1}{5}$.
Hence, $abc+bcd+cde+dea+eab=\frac{1}{25}$ and it's enough to prove that
$$abc+bcd+cde+dea+eab\leq\frac{1}{25}.$$
Let $a=\min\{a,b,c,d,e\}$.
Hence, $0\leq a\leq\frac{1}{5}$ and by AM-GM we obtain
$$abc+bcd+cde+dea+eab=a(b+d)(c+e)+cd(b+e-a)\leq$$
$$\leq a\left(\frac{b+c+d+e}{2}\right)^2+\left(\frac{c+d+b+e-a}{3}\right)^3=\frac{a(1-a)^2}{4}+\frac{(1-2a)^3}{27}$$
and it remains to prove that
$$\frac{a(1-a)^2}{4}+\frac{(1-2a)^3}{27}\leq\frac{1}{25},$$
which is $(5a-1)^2(5a+8)\geq0$ and we are done!
For $n\geq6$ the idea is:
$$F\leq\left(x_1+x_4+...\right)\left(x_2+x_5+...\right)\left(x_3+x_6+...\right)\leq\left(\frac{x_1+x_2+...+x_n}{3}\right)^3=\frac{1}{27}.$$
the equality occurs for $x_1=x_2=x_3=\frac{1}{3},$ which says that the answer is $\frac{1}{27}$.
