Chinese Mathematical Olympiad

The needed constant $C = \frac{1}{2}$. First show that $C = \frac{1}{2}$ holds. Note that
$$\text{L.H.S. of (*)} = x_1^2 + (x_1 + x_2)^2 + \dots + (x_1 + x_2 + \dots + x_n)^2 \\ \qquad + (x_2 + \dots + x_n)^2 + \dots + (x_{n-1} + x_n)^2 + x_n^2. \quad (1)$$
Using the inequality $a^2 + (a+b)^2 = a^2 + (-a-b)^2 \geq \frac{1}{2}b^2$, we get
$$\begin{aligned} \frac{1}{2}(x_1^2 + (x_1 + x_2)^2) &\geq \frac{1}{4}x_2^2, \\ \frac{1}{2}((x_1 + x_2)^2 + (x_1 + x_2 + x_3)^2) &\geq \frac{1}{4}x_3^2, \\ \dots &\geq \dots \\ \frac{1}{2}((x_1 + \dots + x_{n-1})^2 + (x_1 + \dots + x_n)^2) &\geq \frac{1}{4}x_n^2, \\ \frac{1}{2}((x_1 + \dots + x_{n-1})^2 + (x_2 + \dots + x_n)^2) &\geq \frac{1}{4}x_1^2, \\ \dots &\geq \dots \\ \frac{1}{2}((x_{n-1} + x_n)^2 + x_n^2) &\geq \frac{1}{4}x_{n-1}^2. \end{aligned}$$
Summing the above and use (1) to get
$$\text{L.H.S. of (*)} \geq \frac{3}{4}x_1^2 + \frac{1}{2}(x_2^2 + \dots + x_{n-1}^2) + \frac{3}{4}x_n^2 \geq \frac{1}{2}(x_1^2 + \dots + x_n^2).$$
i.e. the inequality (*) holds for $C = \frac{1}{2}$.

Next, we prove that, if the inequality (*) holds for all $n$, then $C \leq \frac{1}{2}$. In (1), take $x_1 = 1, x_2 = -2, x_3 = 2, \dots, x_{n-1} = (-1)^{n-2}2, x_n = (-1)^{n-1}$, then every square in () is equal to 1, i.e. the R.H.S. of () is equal to $2n-2$. Yet
$$\sum_{i=1}^{n} x_i^2 = 4(n-2) + 2 = 4n - 6.$$
So,
$$C \leq \frac{2n-2}{4n-6} = \frac{1}{2} + \frac{1}{2n-3}.$$
Letting $n \to \infty$ gives $C \leq \frac{1}{2}$.

To sum up, the maximal $C = \frac{1}{2}$.