Starting from the basic inequality \(\exp (-x) \geq 1-x\), it's easy to derive by elementary algebraic manipulations the two inequalities

\[

\begin{aligned}

&\exp (x)-x-1 \leq 2(\cosh (x)-1) \\

&2(\cosh (x)-1) \leq x \sinh (x)

\end{aligned}

\]

By the Chernoff-Cramèr bound, we have:

\[

\operatorname{Pr}\left\{\sum_{i=1}^{n}\left(X_{i}-E\left[X_{i}\right]\right)>\varepsilon\right\} \leq \exp \left[-\sup _{t>0}(t \varepsilon-\psi(t))\right]

\]

where

\[

\psi(t)=\sum_{i=1}^{n}\left(\ln E\left[e^{t X_{i}}\right]-t E\left[X_{i}\right]\right)

\]

Keeping in mind that the condition

\[

\operatorname{Pr}\left\{\left|X_{i}\right| \leq M\right\}=1 \quad \forall i

\]

implies that, for all \(i\),

\[

E\left[\left|X_{i}\right|^{k}\right] \leq M^{k} \quad \forall k \geq 0

\]

and since \(\ln x \leq x-1 \forall x>0\), and

\[

E\left[|X|^{k}\right] \leq M^{k} \Longrightarrow E\left[|X|^{k}\right] \leq E\left[X^{2}\right] M^{k-2} \quad \forall k \geq 2, k \in N

\]