How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$? Let 
$$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$
Find the  maximum of the value of
$$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$
where 
$$x_{7}=x_{1},\quad x_{8}=x_{2},\quad x_{9}=x_{3}\,.$$

For $x_i=\frac{1}{6}$ we get $\frac{1}{216}$.
We'll prove that it's a maximal value.
Indeed, let $x_1=\min\{x_i\}$, $x_2=x_1+a$, $x_3=x_1+b$, $x_4=x_1+c$, $x_5=x_1+d$ and $x_6=x_1+e$.
Hence, $a$, $b$, $c$, $d$ and $e$ are non-negatives and we need to prove that:
$$216\sum_{i=1}^6x_ix_{i+1}x_{i+2}x_{i+3}\leq\left(\sum_{i=1}^6x_i\right)^4$$ or
$$216(a^2+b^2+c^2+d^2+e^2-ab-bc-cd-de)x_1^2+$$
$$24((a+b+c+d+e)^3-9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde))x_1+$$
$$+(a+b+c+d+e)^4-216(abcd+bcde),$$
which is true because
$$a^2+b^2+c^2+d^2+e^2-ab-bc-cd-de\geq$$
$$\geq a^2+b^2+c^2+d^2+e^2-ab-bc-cd-de-ea=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0,$$
$$216(abcd+bcde)=216bcd(a+e)\leq216\left(\frac{a+b+c+d+e}{4}\right)^4=$$
$$=\frac{216}{256}(a+b+c+d+e)^4\leq(a+b+c+d+e)^4$$ and
$$(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$$
but my proof of this statement is very ugly.
