Local Mathematical Competitions

First we will prove that $C_n \ge \frac{n-1}{2n}$, i.e. that for any $n$ functions $f_1, f_2, \dots, f_n : [0, 1] \to \mathbb{R}$, there exist numbers $x_1, x_2, \dots, x_n$ in $[0, 1]$ such that
$$|f_1(x_1) + f_2(x_2) + \dots + f_n(x_n) - x_1x_2\dots x_n| \ge \frac{n-1}{2n}.$$
For $n=1$ this is trivial. For $n \ge 2$ suppose, contrariwise, that for all $x_1, x_2, \dots, x_n$ in $[0, 1]$ we have
$$|f_1(x_1) + f_2(x_2) + \dots + f_n(x_n) - x_1x_2\dots x_n| < \frac{n-1}{2n}.$$
Plugging in $x_i = 1$ for $1 \le i \le n$, we get $\left| \sum_{i=1}^{n} f_i(1) - 1 \right| < \frac{n-1}{2n}$.
Plugging in $x_i = 0$ for $1 \le i \le n$, we get $\left| \sum_{i=1}^{n} f_i(0) \right| < \frac{n-1}{2n}$.
Plugging in (for every $1 \le i \le n$) $x_i = 0$ and $x_j = 1$ for all $j \ne i$, we get
$$\left| f_i(0) + \sum_{j \ne i} f_j(1) \right| < \frac{n-1}{2n}.$$
Since
$$(n-1) \sum_{i=1}^{n} f_i(1) = \sum_{i=1}^{n} \left( f_i(0) + \sum_{j \ne i} f_j(1) \right) - \sum_{i=1}^{n} f_i(0),$$
by the triangle inequality we have
$$(n-1) \left| \sum_{i=1}^{n} f_i(1) \right| < (n+1) \frac{n-1}{2n}.$$
On the other hand, by again the triangle inequality we have
$$1 \le \left| \sum_{i=1}^{n} f_i(1) \right| + \left| \sum_{i=1}^{n} f_i(1) - 1 \right| < \frac{n+1}{2n} + \frac{n-1}{2n} = 1,$$
which is a contradiction.

To prove that
$$C_n = \frac{n-1}{2n}$$
is the largest constant, it will be sufficient to prove that for the $n$ (equal) functions
$$f_i(x) = f(x) := \frac{x^n}{n} - \frac{n-1}{2n^2}, \quad 1 \le i \le n,$$
and any $n$ numbers $x_1, x_2, \dots, x_n$ in $[0, 1]$, we have
$$\left| f(x_1) + f(x_2) + \cdots + f(x_n) - x_1 x_2 \cdots x_n \right| \le \frac{n-1}{2n},$$
equivalent to
$$-\frac{n-1}{2n} \le \frac{1}{n} \sum_{i=1}^{n} x_i^n - \prod_{i=1}^{n} x_i - \frac{n-1}{2n} \le \frac{n-1}{2n}.$$
The LHS inequality follows from the AM-GM inequality.

The RHS inequality is equivalent to
$$F(\mathbf{x}) = F(x_1, x_2, \dots, x_n) := \frac{1}{n} \sum_{i=1}^{n} x_i^n - \prod_{i=1}^{n} x_i \le \frac{n-1}{n}$$
at all points $\mathbf{x} = (x_1, x_2, \dots, x_n)$ of the hypercube $[0, 1]^n$. Since $F$ is convex in every variable, its maximum is reached at some vertex $\mathbf{v}$ of the hypercube (point with $x_i = 0$ or $x_i = 1$, for all $1 \le i \le n$). It is easy to see that for all such points we have $F(\mathbf{v}) \le \frac{n-1}{n}$, which completes the proof.