We use the induction for $n \ge 2.$ If $n=2,$ it becomes $$a_1^{2k} \left(a_1 - 1 \right) + a_2^k \left[a_2^k \left(a_2 - 1 \right) - 2a_1^k \right] \ge 0.$$But it is true since $a_2^k \left(a_2 - 1 \right) \ge \left(a_1 + 1 \right)^k a_1 \ge 2a_1^k.$ If the problem is true for $n$, that is $$\sum_{i=1}^n a_i^{2k+1} \ge \left(\sum_{i=1}^n a_i \right)^2,$$we need to show $$ \sum_{i=1}^{n+1} a_i^{2k+1} \ge \left(\sum_{i=1}^{n+1} a_i ^k \right)^2,$$or it is enough to show $$a_{n+1}^{2k+1} + \left(\sum_{i=1}^{n}a_i^k\right)^2 \ge \left(\sum_{i=1}^{n+1} a_i ^k\right)^2,$$$$\Leftrightarrow a_{n+1}^k \left(a_{n+1} - 1 \right) \ge 2 \left(a_1^k + a_2^k + \cdots + a_n^k \right) \ \ \ \ \ (*)$$Again, we prove $(*)$ by the induction for $n \ge 1.$ It is true for $n=1.$ Suppose that $$a_{n}^k \left(a_{n} - 1 \right) \ge 2 \left(a_1^k + a_2^k + \cdots + a_{n-1}^k \right),$$then $$2 \left(a_1^k + a_2^k + \cdots + a_n^k \right) \le a_n^k \left(a_n + 1 \right) \le a_{n+1}^k \left(a_{n+1} - 1 \right),$$because $a_{n+1} \ge a_n+1.$ This completes our proof. The equality holds iff $k=1, \ a_i = i, \ \forall i = \overline{1,n}.$