, 2018

We prove the statement in two steps:

(1) Prove by mathematical induction that:
$$2 \sum_{i=1}^{n} a_i^k \le (a_n + 1)^k a_n. \qquad (1)$$
Proof: When $n=1$, it is easy to see that Eq. (1) holds.
Assume that when $n=m$, Eq. (1) holds, that is
$$2 \sum_{i=1}^{m} a_i^k \le (a_m + 1)^k a_m .$$
Then when $n=m+1$,
$$\begin{aligned} 2 \sum_{i=1}^{m+1} a_i^k &= 2 \sum_{i=1}^{m} a_i^k + 2a_{m+1}^k \\ &\le (a_m + 1)^k a_m + 2a_{m+1}^k \\ &\le a_{m+1}^k (a_{m+1} - 1) + 2a_{m+1}^k \\ &= a_{m+1}^k (a_{m+1} + 1) \\ &\le (a_{m+1} + 1)^k a_{m+1}. \end{aligned}$$

(2) Next, prove that the statement holds by mathematical induction.
When $n=1$, it is easy to see that the statement holds.
Assume that when $n=m$, the statement holds, that is
$$\left( \sum_{i=1}^{m} a_i^k \right)^2 \le \sum_{i=1}^{m} a_i^{2k+1} .$$

$$\begin{aligned} & \left( \sum_{i=1}^{m+1} a_i^k \right)^2 \\ &= \left( \sum_{i=1}^{m} a_i^k \right)^2 + 2 \left( \sum_{i=1}^{m} a_i^k \right) a_{m+1}^k + a_{m+1}^{2k} \\ &\le \sum_{i=1}^{m} a_i^{2k+1} + 2 \left( \sum_{i=1}^{m} a_i^k \right) a_{m+1}^k + a_{m+1}^{2k} \\ &\le \sum_{i=1}^{m} a_i^{2k+1} + (a_m + 1)^k a_m a_{m+1}^k + a_{m+1}^{2k} \quad \text{(by Eq. (1))} \\ &\le \sum_{i=1}^{m} a_i^{2k+1} + a_{m+1}^k \left( a_{m+1}^k (a_{m+1}-1) + a_{m+1}^k \right) \\ &\le \sum_{i=1}^{m+1} a_i^{2k+1}. \end{aligned}$$

Combining the above, the statement holds.