Chinese Mathematical Olympiad

The needed constant $C=\frac{1}{2}$. First show that $C=\frac{1}{2}$ holds. Note that $$\begin{gathered} \text{L.H.S. of (*)}=x_1^2+(x_1+x_2{)}^2+\cdots +(x_1+x_2+\cdots +x_n{)}^2 \\ +(x_2+\cdots +x_n{)}^2+\cdots +(x_{n-1}+x_n{)}^2+x_n^2.(1) \end{gathered}$$ Using the inequality $a^2+(a+b{)}^2=a^2+(-a-b{)}^2\ge \frac{1}{2}{b}^2$, we get $$\begin{array}{rl}\frac{1}{2}(x_1^2+(x_1+x_2{)}^2)& \ge \frac{1}{4}{x}_2^2,\\ \frac{1}{2}((x_1+x_2{)}^2+(x_1+x_2+x_3{)}^2)& \ge \frac{1}{4}{x}_3^2,\\ \dots & \ge \dots \\ \frac{1}{2}((x_1+\cdots +x_{n-1}{)}^2+(x_1+\cdots +x_n{)}^2)& \ge \frac{1}{4}{x}_n^2,\\ \frac{1}{2}((x_1+\cdots +x_{n-1}{)}^2+(x_2+\cdots +x_n{)}^2)& \ge \frac{1}{4}{x}_1^2,\\ \dots & \ge \dots \\ \frac{1}{2}((x_{n-1}+x_n{)}^2+x_n^2)& \ge \frac{1}{4}{x}_{n-1}^2.\end{array}$$ Summing the above and use (1) to get $$\text{L.H.S. of (*)}\ge \frac{3}{4}{x}_1^2+\frac{1}{2}(x_2^2+\cdots +x_{n-1}^2)+\frac{3}{4}{x}_n^2\ge \frac{1}{2}(x_1^2+\cdots +x_n^2).$$ i.e. the inequality () holds for $C=\frac{1}{2}$.

Next, we prove that, if the inequality () holds for all $n$, then $C\le \frac{1}{2}$. 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, $$C\le \frac{2n-2}{4n-6}=\frac{1}{2}+\frac{1}{2n-3}.$$ Letting $n\to \infty$ gives $C\le \frac{1}{2}$.

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