prove this nice inequality $\left|\prod_{i=1}^{n}(a_{i}-a_{i+1})\right|\le \frac{3\sqrt{3}}{16}$ Let $n$ be odd number, and $a_{i}\ge 0$, such that
$$2(a_{1}+a_{2}+\cdots+a_{n})=n$$
Show that
$$\left|\prod_{i=1}^{n}(a_{i}-a_{i+1})\right|\le \frac{3\sqrt{3}}{16}$$
where $a_{n+1}=a_{1}$
Seeing this inequality reminds me to use this conclusion to deal with it.
$$S_{AOB}=\frac{1}{2}|a_{1}b_{2}-a_{2}b_{1}|$$
where $A(a_{1},b_{1}),B(a_{2},b_{2}),O(0,0)$
It seems that this problem will involve a problem of maximum geometric area, perhaps it can be handled this way
 A: Here is a proof for $n=3$, perhaps that helps someone to solve the
general problem.

Let $a, b, c$ be non-negative real numbers with $a+b+c = \frac 32$.
  Then
  $$ f(a, b, c) =  \vert (a-b)(b-c)(c-a) \vert \le \frac{3 \sqrt 3}{16} \, .
$$ 
  Equality holds if and only if $(a, b, c)$ is a permutation of
  $$
 (0,  \frac{3-\sqrt 3}{4}, \frac{3+\sqrt 3}{4} ) \, .
$$

Proof: Without loss of generality we can assume that $a \le b \le c$.
If $a > 0$ then we can replace $(a, b, c)$ with
$$
 (a', b', c') = (0, b + \frac\delta 2, c + \frac\delta 2)
$$
where $\delta = \frac 32 - b - c$. Then $a'+b'+c' = \frac 32$ and
$$
 f(a, b, c) = (b-a)(c-b)(c-a) < b'(c'-b')c' = f(a', b', c') \, .
$$
Therefore we can assume that $a=0$ and it remains to 
maximize $f(0, b, c)$
for $0 \le b \le c, b+c = \frac 32$. It is convenient to set
$$
   b = \frac 34 - x \, , \quad c = \frac 34 + x \, , \quad 0 \le x \le \frac 34 \, .
$$
Then
$$
 f(0, b, c) = b (c-b) c = 2 x \left( \frac {9}{16} - x^2 \right) =: \phi(x) \, .
$$ 
An elementary analysis shows that  $\phi$ has a unique maximum on $[0, \frac 34]$, it is attained at $x_0 = \frac{\sqrt 3}{4}$, and
$\phi(x_0) = \frac{3 \sqrt 3}{16}$. 
This concludes the proof.

Remark: For arbitrary odd $n \ge 3$ the upper bound
 $\frac{3 \sqrt 3}{16}$ is best possible. For $n=3$ that has been
demonstrated above. For odd $n \ge 5$ equality holds for
$$
 (a_1, \ldots, a_n) = (0,  \frac{3-\sqrt 3}{4}, \frac{3+\sqrt 3}{4}, 0, 1, \ldots, 0, 1) \, .
$$
