How find this nice minmum of this value $\frac{\prod_{i=1}^{n-1}(a_{i}+a_{i+1})\sum a_{i}}{\prod_{i=1}^{n}a_{i}}$ Let $n$ be give postive integer number. For any $a_{i}>0 (i=1,2,\cdots,n)$, find the minimum of the value
$$F_{n}(a_{1},a_{2},\cdots,a_{n})=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})\cdots (a_{n-1}+a_{n})(a_{1}+a_{2}+\cdots+a_{n})}{a_{1}a_{2}a_{3}\cdots a_{n}}.$$(by wang yong xi)
I try when $n=2$,then
$$F_{2}(a_{1},a_{2})=\dfrac{(a_{1}+a_{2})(a_{1}+a_{2})}{a_{1}a_{2}}=\dfrac{(a_{1}+a_{2})^2}{a_{1}a_{2}}\ge 4$$when $a_{1}=a_{2}$  is minimum
(2):when $n=3$,
$$F_{3}=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})(a_{1}+a_{2}+a_{3})}{a_{1}a_{2}a_{3}}$$WLOG $a_{3}=1$,so
$$F=\dfrac{(a_{1}+a_{2})(a_{2}+1)(a_{1}+a_{2}+1)}{a_{1}a_{2}}$$
I use this find this minimum is $$(F_{3})_{min}=\dfrac{1}{2}(11+5\sqrt{5})$$
when $a_{1}=1,a_{2}=\dfrac{\sqrt{5}-1}{2}$ seelinks

Here's some partial progress. Maybe someone will have further ideas.
By homogeneity, we can set $a_1 + \cdots + a_n = 1$ throughout. First dispense with boundary behavior. Fix $1 \leq k \leq n$ and note that
$$\frac{\prod_{i=1}^k (a_i+a_{i+1})}{\prod_{i=1}^n a_i} = \prod_{i=1}^{k-1} \frac{a_i + a_{i-1}}{a_i} \cdot \prod_{i=k}^{n-1} \frac{a_i + a_{i+1}}{a_{i+1}} \cdot \frac{1}{a_k} \geq \frac{1}{a_k}.$$
Hence $F_n(a_1, \ldots, a_n) \geq \frac{1}{\min_i a_i}$, so $F_n \to \infty$ uniformly as we tend to the boundary.
Let $G_n(a_1, \ldots, a_n) = \log F_n(a_1, \ldots, a_n)$, which we may as well minimize. Recalling that $a_1+\cdots+a_n=1$ throughout, consider when $\nabla G_n = 0$. We find
$$\begin{align*}a_1 \partial_1 G_n &= \frac{a_1}{a_1+a_2} + a_1 - 1 \\
a_i \partial_i G_n &= \frac{a_i}{a_{i-1} + a_i} + \frac{a_i}{a_i + a_{i+1}} + a_i - 1 \qquad \text{for }1 < i < n \tag{*}\label{*}\\
a_n \partial_n G_n &= \frac{a_n}{a_{n-1}+a_n} + a_n - 1.
\end{align*}$$
In particular we have the relation
$$\sum_{i=1}^n a_i \partial_i G_n = 0$$
which seems important, though I haven't found a use for it.
Using $\eqref{*}$, we can rewrite the condition $\nabla G_n = 0$ as either
$$\begin{align*}
\frac{a_2}{a_1+a_2} &= a_1 \\
\frac{a_3}{a_2+a_3} &= a_1 + a_2 \\
&\vdots \\
\frac{a_n}{a_{n-1}+a_n} &= a_1 + a_2 + \cdots + a_{n-1} \\
1 &= a_1+\cdots+a_n
\end{align*}$$
or
$$\begin{align*}
\frac{a_2}{a_1+a_2} &= a_1 \\
\frac{a_3}{a_2+a_3} &= \frac{a_2}{a_1+a_2} + a_2 \\
&\vdots \\
\frac{a_n}{a_{n-1}+a_n} &= \frac{a_{n-1}}{a_{n-2}+a_{n-1}} + a_{n-1} \\
a_1+\cdots+a_n &= 1.
\end{align*}$$
Either of these forms is suitable for computer algebra systems. I threw the first 12 into Mathematica; here's the output. Using the first $n-1$ relations, we may iteratively solve for $a_2, a_3, \ldots, a_n$ as rational functions of $a_1$, and then we may impose the final condition to solve for $a_1$. This shows that $a_1$ (and hence all the others) are algebraic, so the algebraic minimum math110 suggests in a comment above is at least plausible on its face. Edit: indeed, I checked through $n=12$ that the values of $F_n$ when $\nabla G_n = 0$ are precisely the Galois conjugates of math110's guess of $\left(\sin\left(\frac{2\pi}{n+2}\right)/\sin\left(\frac{\pi}{n+2}\right)\right)^{n+2}$.
Based on that data, it seems that when $n$ is odd, $a_1$ satisfies $p_n(t) := \sum_{k=0}^{(n+1)/2} \binom{n+1-k}{k} (-t)^k = 0$. Intriguingly, this family of polynomials appears to be tending pointwise to 0 on $[0, 1]$ as $n \to \infty$, and they seem to be real-rooted. Moreover, it appears that the unique value of $a_1$ solving the above system subject to $a_i > 0$ and $a_1 + \cdots + a_n = 1$ is the smallest root of $p_n(t)$.
I get the sense that I'm missing some sort of physical meaning in this problem. I also can't see why $\eqref{*}$ implies the symmetry $a_i = a_{n-i+1}$, though it should. There's a bizarre amount of structure throughout this problem that I'm at a loss to explain!
