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You can prove it by using the fact that the following holds always:

$\max_{1 \le k \le n}|S_k| \le \max_{1 \le k \le n/2}|S_k| + \max_{1 \le k \le n/2}|M_k|$

If the left hand side is larger than $\epsilon$ then one of the right hand terms is larger than $\epsilon/2$.

This also shows that the inequality is valid under absolutely no assumptions on the joint distribution of the variables $X_1,\ldots,X_n$.

1

You can prove it by using the fact that the following holds always:

$\max_{1 \le k \le n}|S_k| \le \max_{1 \le k \le n/2}|S_k| + \max_{1 \le k \le n/2}|M_k|$

If the left hand is larger than $\epsilon$ then one of the right hand terms is larger than $\epsilon/2$.

This also shows that the inequality is valid under absolutely no assumptions on the joint distribution of the variables $X_1,\ldots,X_n$.