I was interested in the following problem regarding a certain combination of 7 real numbers that sum up to 1, the problem states:
Given $x_1, x_2, x_3,...x_7\in \Bbb R_{\ge 0}$ such that $x_1+x_2+x_3+...+x_7=1$, what is the maximum of $x_1x_2+x_2x_3+x_3x_4+x_4x_5+x_5x_6+x_6x_7+x_7x_1$?
At first I tried to solve an easier version of this problem with a smaller number of terms in order to get some insight for the original problem. For example in the case with 2 real numbers s.t. $x_1+x_2=1$ and $x_1x_2+x_2x_1$ is almost trivial and can be dealt with by a simple substitution $x_1=1-x_2$, which gives us a maximum of $\frac{1}{2}$. A similar strategy can be used in the case with 4 terms, where the expression factors as $(x_1+x_3)(x_2+x_4)$ and the maximum turns out to be $\frac{1}{4}$. I suppose that the desired maximum is also $\frac{1}{4}$, in fact I was able to prove that for any even $n$ greater than 2, where $n$ is a number of terms, $\frac{1}{4}$ is indeed the maximum.
Let's label $M_n(x_i)=x_1x_2+x_2x_3+x_3x_4+...+x_nx_1$, than for $k\ge2$: $$\max(M_{2k}(x_i))=\frac{1}{4}$$ Proof:
First important observation, is that in the case of $n=2k$ every summand in $M_{2k}(x_i)$ is of the form $x_{2m-1}x_{2m}$ for some $1\le m \le k$. This is not the case when $n=2k-1$ since the last term in $M_{2k-1}(x_i)$ is $x_{2k-1}x_1$ where the indexes have the same parity and so not all summands are of the form $x_{2m-1}x_{2m}$.
The next observation is that the product $(x_2+x_4+x_6+...+x_{2k})(x_1+x_3+x_5+...+x_{2k-1})$ is greater than, or equal to $M_{2k}(x_i)$. This is obvious, since in this product every summand in $M_{2k}(x_i)$ will appear, but it will also include a bunch of other non-negative terms. $$(x_2+x_4+x_6+...+x_{2k})(x_1+x_3+x_5+...+x_{2k-1})=M_{2k}(x_i)+{positive}$$ $$M_{2k}(x_i) \le (x_2+x_4+x_6+...+x_{2k})(x_1+x_3+x_5+...+x_{2k-1})$$ Now, let's call $(x_2+x_4+x_6+...+x_{2k})=s$, then $(x_1+x_3+x_5+...+x_{2k-1})=1-s$. Our inequality turns into $$M_{2k}(x_i)\le s(1-s)$$ We can easily show that the maximum of $s(1-s)$ for $s \in (0; 1)$ is in fact $\frac{1}{4}$, for example by differentiating $s(1-s)$ with respect to $s$.
It suffices to show that $M_{2k}(x_i)$ actually equals to $\frac{1}{4}$ for some $x_1, x_2, ..., x_{2k}$. Take $x_1=x_2=\frac{1}{2}$ and all of the other terms set to $0$, in this case, the sum of $x_i$'s is 1, and $M_{2k}(x_i)=\frac{1}{4}$.
I thought that I could use the same approach for $n=7$ but the essential inequality $$M_{7}(x_i) \le (x_2+x_4+x_6)(x_1+x_3+x_5+x_7)$$ doesn't hold in general, because the product on the right side doesn't include the term $x_7x_1$. We can take $x_2=x_3=x_4=x_5=x_6=0$ and $x_1=x_7=\frac{1}{2}$ and we get that $\frac{1}{4}\le 0$, which is absurd.
I think that I am moving in the right direction, but sadly can not think of any further steps of how to solve it, so I am asking You to help me out.
