$C(n)=\frac{1}{n-1}$ with equality when all $x_{i}=\frac{1}{2}$ Observe that changing $x-e$,$x+e$ into $x$, $x$ doesn't change LHS while it increases RHS. Now if for all $i$: $x_{i}\leq \frac{1}{2}$, we're done (. So assume that $x_{1}=x_{2}=...=x_{n-1}\leq \frac{1}{2}\leq x_{n}$ (there is at most one number greater than $\frac{1}{2}$, because of the condition on $x_{i},x_{j}$). Using $(1-x_{n})(1-x_{1})\geq \frac{1}{4}$ we conclude that $\sum x_{i}\leq \frac{n}{2}$ and so we can assume all variables are equal because $(1-\frac{\sum x_{i}}{n})^{2}\geq \frac{1}{4}$ and hence the maximum is attained with all $x_{i}$ being equal to $\frac{1}{2}$.

Well, I don't understand why you had to go through all that. You've proved that, for a fixed value of $\sum x_{i}$, if there are still unequal terms, you can change it so that the right side gets bigger and everything still satisfies the condition. So the maximum of the right side occurs when all $x_{i}$ are equal...

Because we want $(1-x_{i})(1-x_{j})\geq \frac{1}{4}$ to hold for new variables as well and my later considerations were just to check that.

Well，I think that this prove has a few drawbacks. The adjustment is about two variables x_i and x_j. So you need to confirm that to any i and j，(x_i+x_j)/2 <1/2. Even though you get that the sum of all x_i <n/2, you cannot do such an adjustment for x_n and x_i (consider the instance of (0.0001,0.0001,...,0.0001,1). Meanwhile, l cannot come across how you get the sum of all x_i <n/2, and with all x_i equaling to one value, assumed as x, you need to confirm that when x equals to 1/2, the maximum of C(n) occurs（while this step is easy）．