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
$$a_k = a_k,$$
$$a_{k+1} = a_k - d_k,\quad a_{k+2} = a_k - d_k - d_{k+1}, \dots,$$
$$a_n = a_k - d_k - d_{k+1} - \dots - d_{n-1},$$
$$a_{k-1} = a_k + d_{k-1},\quad a_{k-2} = a_k + d_{k-1} + d_{k-2}, \dots,$$
$$a_1 = a_k + d_{k-1} + d_{k-2} + \dots + d_1.$$
Summing all equalities, and using $a_1 + a_2 + \dots + a_n = 0$, we get
$$na_k - (n-k)d_k - (n-k-1)d_{k+1} - \dots - d_{n-1} \\ \quad + (k-1)d_{k-1} + (k-2)d_{k-2} + \dots + d_1 = 0.$$
Then in view of Cauchy's inequality
$$\begin{aligned} (na_k)^2 &= ((n-k)d_k + (n-k-1)d_{k+1} + \dots + d_{n-1} \\ &\quad - (k-1)d_{k-1} - (k-2)d_{k-2} - \dots - d_1)^2 \\ &\le \left[ \sum_{i=1}^{k-1} i^2 + \sum_{i=1}^{n-k} i^2 \right] \left[ \sum_{i=1}^{n-1} d_i^2 \right] \\ &\le \left[ \sum_{i=1}^{n-1} i^2 \right] \left[ \sum_{i=1}^{n-1} d_i^2 \right] = \frac{n(n-1)(2n-1)}{6} \left[ \sum_{i=1}^{n-1} d_i^2 \right] \\ &\le \frac{n^3}{3} \left[ \sum_{i=1}^{n-1} d_i^2 \right]. \end{aligned}$$
So, $a_k^2 \le \frac{n}{3} \sum_{i=1}^{n-1} (a_i - a_{i+1})^2$.