Kind of maximal inequality. I really could need some external impressions on the following problem:
Consider a sequence of random variables $(X_j)_{j \in \mathbb{N}}$, which is bounded, stationary and even $m$-dependent for an $m \in \mathbb{N}$. Now let $k \in \mathbb{N}$ such that $2^k > m$. Then I want to prove that for $x > 0$ the following (maximal) inequality holds true:

$$\tag{A}\mathbb{P}(\max_{1\leq l \leq 2^k}\vert \sum_{n=1}^l X_n - \sum_{j=1}^{\lfloor l/2m\rfloor} \sum_{n=1+2jm}^{2(j+1)m} X_n \vert > x) \leq \frac{2^k}{m} \mathbb{P}(\max_{1\leq l \leq 2m} \vert \sum_{i=1}^l X_i \vert > x)$$

Yep. Looks ugly. I have tried to find a suitable maximal inequality, something like the Levy inequality, but I wasn't successful. Further I tried it without using any external results and just use the basic properties of probability measures, but I couldn't figure it out.
I was able to compute the expression $$\sum_{n=1}^l X_n - \sum_{j=1}^{\lfloor l/2m\rfloor} \sum_{n=1+2jm}^{2(j+1)m} X_n = \underbrace{\sum_{n=1}^{2m}X_n}_{I} - \underbrace{\sum_{j=1+l}^{2(\lfloor l/2m \rfloor + 1)m} X_j}_{II} .$$
By the stationarity I think, that both $I$ and $II$ respectively should satisfy $$ \mathbb{P}(\vert I \vert > x), \mathbb{P}(\vert II \vert > x) \leq \mathbb{P}(\max_{1\leq l \leq 2m} \vert \sum_{i=1}^l X_i \vert > x).$$
Although this made me feel like that was a step in the right direction, I still have no idea how to proceed. Especially the fraction $2^k/m$ makes me feel, that only the cases where $l$ is a multiple of $m$ should madder for the maximum on the righthandside in (A), but I probably have spent to many time and can not think clear anymore.
Any suggestions for literature, known theorems, ideas, hints is very much appreciated.

First notice that
$$
\left\lvert\sum_{n=1}^\ell X_n - \sum_{j=0}^{\lfloor \ell/2m\rfloor} \sum_{n=1+2jm}^{2(j+1)m} X_n\right\rvert =\left\lvert\sum^{2(\lfloor \ell/2m\rfloor+1)m+1}_{n=\ell}
X_n\right\rvert .$$
Express $\ell$ as $2mq(\ell)+r(\ell)$, where $0\leqslant r(\ell)\leqslant 2m-1$. Then
$$
\left\lvert \sum_{n=1}^\ell X_n - \sum_{j=0}^{\lfloor \ell/2m\rfloor} \sum_{n=1+2jm}^{2(j+1)m} X_n\right\rvert =\left\lvert \sum^{2(q(\ell)+1)m+1}_{n=2mq(\ell)+r(\ell)}
X_n\right\rvert \leqslant \max_{0\leqslant q\leqslant \lfloor 2^{k-1}/m\rfloor+1 }\max_{0\leqslant r\leqslant 2m-1}\left\lvert \sum^{2(q +1)m+1}_{n=2mq +r }
X_n\right\rvert .
$$Then use a union bound and stationarity.
