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
The general strategy is to show that, without loss of generality, the $a_i$'s must be of a specified form. If not, we could replace the $a_i$ by another sequence with the same sum but equal or larger cyclic product of absolute differences. All of our replacements below are of this type. [We include a running example in brackets.]
Then it will be enough to maximize that product over all sequences of the specified form.
[Ex: .85, .98, .18, .34, .21, .74, .92, .20, .08]
Wlog, the smallest $a_i$ is zero. If not, let $b=\min(a_i)$. We can replace one minimal $a_i$ with $0$, and replace each of the other $a_i$ by $a_i+\frac{b}{n-1}$.
[Ex: .86, .99, .19, .35, .22, .75, .93, .21, .00]
Wlog, any three consecutive $a_i$ include at least one zero. If not, suppose that the $a_i$ contain the subsequence $0,b,c,d,e$ with $b,c,d$ positive. Let $m$ be the average of $b,c,d$.
[Ex: .955, 1.085, .00, .35, .22, .75, .93, .21, .00]
[Ex: .955, 1.085, .00, .57, .00, .75, .21, .93, .00]
[Ex: .955, 1.085, .00, .57, .00, .96, .00, .93, .00]
Mathematica can verify quickly that these replacements increase the product:
f[x_, y_, z_] := Abs[x (y - x) (z - y)]
positive[s_] := {0 < b, 0 < c, 0 < d, m == (b + c + d)/3, s}
Simplify[f[b, c, d] <= f[b + c, 0, d], Assumptions -> positive[d > m]]
Simplify[f[b, c, d] <= f[2 m, 0, m], Assumptions -> positive[d < m]]
Simplify[f[b, c, d] <= f[b + c, 0, d], Assumptions -> positive[d > c]]
Simplify[f[b, c, d] <= f[b + d/2, c + d/2, 0], Assumptions -> positive[d < c]]
(WolframAlpha provides the same verifications here, here, here and here, by saying "no solutions exist" to a search for counterexamples; this uses a modern variant of Tarski's algorithm for determining when polynomials have solutions in the reals.)
Wlog, any three consecutive $a_i$ include at least one zero; also there is at most one triplet in $a_i$ with signs $0,+,+$. If not, and the triplets are not next to each other, we can replace the subsequence $0,a,b,\ldots,0,c,d,\ldots$ with its rearrangement $0,a,b,0,c,d,\ldots,\ldots$. Then if we have the subsequence $0,b,c,0,d,e,0$, we can assume wlog that $b>c$ and $d>e$, and can replace the subsequence with $0,b,0,d,0,c+e,0$.
[Ex: .955, 1.085, .00, .57, .00, .96, .00, .93, .00]
Wlog, any three consecutive $a_i$ include at least one zero; also, whenever zero is followed by a single positive number, it is always the same positive number. If not, we can replace $0,b,0,c,0$ by $0,(b+c)/2,0,(b+c)/2,0$, and similarly with means of longer sequences.
[Ex: .955, 1.085, .00, .82, .00, .82, .00, .82, .00]
So wlog, the sequence is of the form $0,b,c,0,d,0,d,\ldots,0,d$ with $b>c$.
[Ex: .00, 1.085, .955, .00, .82, .00, .82, .00, .82]
The sum condition gives $d=(n-2b-2c)/(n-3)$. This gives the product $$b(b-c)c\left(\frac{n-2b-2c}{n-3}\right)^{n-3}$$ and we can maximize this by maximizing $$\ln(b)+\ln(b-c)+\ln(c)+(n-3)\ln(n-2b-2c).$$ Setting partial derivatives equal to $0$ gives $$\frac{1}{b}+\frac{1}{b-c}+\frac{-2(n-3)}{n-2b-2c}=0$$ $$\frac{-1}{b-c}+\frac{1}{c}+\frac{-2(n-3)}{n-2b-2c}=0$$ Solving with the constraint $b>c$, $n>0$, we get $$b=\frac{3+\sqrt{3}}{4},\ c = \frac{3-\sqrt{3}}{4}$$ which leads to the maximal sequence $$0, \frac{3+\sqrt{3}}{4},\frac{3-\sqrt{3}}{4},0,1,0,1,\ldots,0,1$$ and its cyclic product of absolute differences $$\frac{3\sqrt{3}}{16}.$$
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 $$ (a_1, \ldots, a_n) = (0, \frac{3-\sqrt 3}{4}, \frac{3+\sqrt 3}{4}, 0, 1, \ldots, 0, 1) \, . $$
