China Mathematical Olympiad

It is sufficient to prove that: For every $k\in \{1,2,\dots ,n\}$, we have $$a_k^2\le \frac{n}{3}\sum _{i=1}^{n-1}(a_i-a_{i+1}{)}^2.$$ Let $d_k=a_k-a_{k+1}$, $k=1,2,\dots ,n-1$, then $$a_k=a_k,$$ $$a_{k+1}=a_k-d_k,a_{k+2}=a_k-d_k-d_{k+1},\dots ,$$ $$a_n=a_k-d_k-d_{k+1}-\cdots -d_{n-1},$$ $$a_{k-1}=a_k+d_{k-1},a_{k-2}=a_k+d_{k-1}+d_{k-2},\dots ,$$ $$a_1=a_k+d_{k-1}+d_{k-2}+\cdots +d_1.$$ Summing all equalities, and using $a_1+a_2+\cdots +a_n=0$, we get $$\begin{gathered} n{a}_k-(n-k)d_k-(n-k-1)d_{k+1}-\cdots -d_{n-1} \\ +(k-1)d_{k-1}+(k-2)d_{k-2}+\cdots +d_1=0. \end{gathered}$$ Then in view of Cauchy's inequality $$\begin{array}{rl}(n{a}_k{)}^2& =((n-k)d_k+(n-k-1)d_{k+1}+\cdots +d_{n-1}\\ & -(k-1)d_{k-1}-(k-2)d_{k-2}-\cdots -d_1{)}^2\\ & \le \left[\sum _{i=1}^{k-1}{i}^2+\sum _{i=1}^{n-k}{i}^2\right]\left[\sum _{i=1}^{n-1}{d}_i^2\right]\\ & \le \left[\sum _{i=1}^{n-1}{i}^2\right]\left[\sum _{i=1}^{n-1}{d}_i^2\right]=\frac{n(n-1)(2n-1)}{6}\left[\sum _{i=1}^{n-1}{d}_i^2\right]\\ & \le \frac{n^3}{3}\left[\sum _{i=1}^{n-1}{d}_i^2\right].\end{array}$$ So, $a_k^2\le \frac{n}{3}{\sum }_{i=1}^{n-1}(a_i-a_{i+1}{)}^2$.