Minimization of a cyclic sum with a certain function

Let $f:[0,2]^2 \rightarrow [0,1]$ be defined as $$f(x,y):=-\dfrac{x}{4} \sqrt{4-x^2}-\dfrac {1}{4 \sqrt 2}\sqrt{2(x^2+y^2)-x^2y^2+xy \sqrt{(4-x^2)(4-y^2)}}+\sin \dfrac {\arccos{\dfrac{xy-\sqrt{(4-x^2)(4-y^2)}}{4}}+\arccos{ \left (1-\dfrac {x^2}{2} \right )}}{2}.$$ Suppose we have $n$ variables $0 \leq x_i \leq 2$ ($1 \leq i \leq n$) such that the sum $\sum_{i=1} ^n x_i=S$ is fixed. What is the least value of $$\Sigma=\sum _{i=1} ^n f(x_i, x_{i+1}) \quad (x_{n+1}=x_1)$$ over $[0,2]^n$? Is it true that we can attain the minimum for $x_1=x_2=\dots =x_n$? If not, what is this minimum? Lagrange multipliers are out of reach here because of the complexity of $f$. Is there a no-calculations proof to find this minimum, or otherwise can we use a CAS to find it? I tried to calculate $\dfrac {\partial \Sigma}{ \partial x_i}=\partial _y f(x_{i-1}, x_i)+\partial _x f(x_i, x_{i+1})$ then set it equal to $0$, but that didn't lead to any result.

A little more detail: it may seem manageable to minimize $f(x,y)+f(y,z)$ as a function of $y$, or use a similar approach, but I couldn't still find computer algebra systems which minimize a function of three variables-or how to do it.

It may be that the minimum is effectively not attained on the line $x_1= \dots =x_n$, but in this case I'd need to prove that the "true" global minimum is within a distance $o(n)$ of the minimum on this line.

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Is this homework from some mean teacher? – 1015 Mar 23 '13 at 14:54
Unfortunately no, this came out from a research problem I'm working on. Faced with this kind of problem I had always just rushed through Lagrange multipliers, but here it seems to be just impossible (I hope someone will refute this...)! – lookatthis Mar 23 '13 at 15:00

I have solved it today. It is simpler than it seemed: for the function $f$ that I wrote, just break the $-\frac x 4 \sqrt{4-x^2}$ part in two parts and shift each half to get a function that is almost symmetric and whose cyclic sum we can minimize up to an error term of $\varepsilon <10^{-10}$, which is enough for my purposes.