Let $\epsilon > 0, \, (X_i)_{i\geq1}$ be a sequence of identically distributed (we denote $EX_i$ by $\mu$) and pairwise (not mutually!) independent random variables
prove that there exists some constants $C$ such that:
$\mathbb{P}[|\sum_{i=1}^{n}(X_i - \mu)| > \epsilon n] \leq C(\sum_{i=1}^{n}\mathbb{P}[|X_i|>n]+\mathbb{P}[|\sum_{i=1}^n(X_i-\mu)\mathbb{I}_{\{|X_i| \leq n\}}|>\frac{\epsilon n}{2}])$
this inequality was used in the second part of a math paper I was reading, I've been trying to prove it for the last couple of days but to no avail, if anyone could point me in the right direction that would be great.
Write $$\begin{multline*}\mathbb{P}\left(\left|\sum_{i=1}^n (X_i-\mu) \right|>\epsilon n\right) = \mathbb{P}\left(\left|\sum_{i=1}^n (X_i-\mu) \right|>\epsilon n, \, |X_i| >n \text{ for some } i\right) \\+ \mathbb{P}\left(\left|\sum_{i=1}^n (X_i-\mu) \right|>\epsilon n, \, |X_i|\leq n \text{ for every } i\right).\end{multline*}$$ The first term is less than $\mathbb{P}(|X_i| > n \text{ for some } i) \leq \sum_{i=1}^n \mathbb{P}(|X_i|>n)$. The second term is equal to $$\mathbb{P}\left(\left|\sum_{i=1}^n (X_i-\mu) \mathbf{1}_{|X_i|\leq n}\right|>\epsilon n, \, |X_i|\leq n \text{ for every } i\right) \leq \mathbb{P}\left(\left|\sum_{i=1}^n (X_i-\mu) \mathbf{1}_{|X_i|\leq n}\right|>\epsilon n\right)$$ which proves your inequality.