, 2011

Make the substitution $a_k = b_k + k$ for $k = 1, 2, \dots, n$. The conditions of the problem then become $b_m + b_{m+1} + \dots + b_n \ge 0$ for every $m = 1, 2, \dots, n$, while the required inequality becomes
$$\begin{aligned} \frac{n(n+1)(2n+1)}{6} &\le (b_1+1)^2 + (b_2+2)^2 + \dots + (b_n+n)^2 \\ &= (b_1^2 + b_2^2 + \dots + b_n^2) + (1^2 + 2^2 + \dots + n^2) + 2(b_1 + 2b_2 + \dots + n b_n). \end{aligned}$$
But $b_1^2 + \dots + b_n^2 \ge 0$, $1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$ while
$b_1+2b_2+\dots+nb_n = b_n+(b_{n-1}+b_n)+(b_{n-2}+b_{n-1}+b_n)+\dots+(b_1+b_2+\dots+b_n) \ge 0$
by the (restated) given condition.