Bernstein-Chernoff inequality I am studying the proof of the following result due to Bernstein and Chernoff:
Let $\xi_1,\ldots,\xi_m$ be independent random variables with $|\xi_i|\leq 1$ for $i=1,\ldots,m$. Let $\beta = \sum_{i=1}^m\mathbb{E}[(\xi_i-\mathbb{E}[\xi_i])^2]$. Then 
$$\mathbb{P}\bigg(\bigg|\sum_{i=1}^m \big(\xi_i-\mathbb{E}[\xi_i]\big)\bigg|\geq \gamma\bigg)\leq 2\exp\left(-\frac{\gamma}{4}\right),\text{ if }\gamma \geq \beta,$$ and $$\mathbb{P}\bigg(\bigg|\sum_{i=1}^m \big(\xi_i-\mathbb{E}[\xi_i]\big)\bigg|\geq \gamma\bigg)\leq 2\exp\left(-\frac{\gamma^2}{4\beta}\right),\text{ if }\gamma \leq \beta.$$
At some point, we have to obtain an upper bound of $\mathbb{E}[e^{y\xi_i}]$, with $0<y\leq 1$. To obtain this, the author uses exponential's series expansion, and then claims that the following inequality holds: $$\mathbb{E}[e^{y\xi_i}]\leq 1+\frac{y^2\mathbb{E}[\xi_i^2]}{2}+\frac{y^3\mathbb{E}[\xi_i^2]}{6-2y}.$$ I understand that, without loss of generality can assume that the random variables satisfy $\mathbb{E}[\xi_i]=0$, but this only justifies that the term with the first-moment disappears. Any ideas or possible proofs?
 A: \begin{align}
\mathbb{E}\left[e^{y\xi_i}\right] &= \mathbb E\left[\sum_{n \ge 0} \frac{y^n\xi_i^n}{n!}\right]\\
&= 1 + \mathbb E\left[y\xi_i\right] + \frac12\mathbb E\left[y^2\xi_i^2\right] + \mathbb E\left[\sum_{n \ge 3} \frac{y^n \xi_i^n}{n!}\right]\\
&\underset{}{\le} 1 + \mathbb E\left[y\xi_i\right] + \frac12\mathbb E\left[y^2\xi_i^2\right] + \mathbb E\left[\sum_{n \ge 3} \frac{y^n \left|\xi_i\right|^n}{n!}\right]\\
&\underset{|\xi_i| \le 1}{\le} 1 + \mathbb E\left[y\xi_i\right] + \frac12\mathbb E\left[y^2\xi_i^2\right] + \mathbb E\left[\sum_{n \ge 3} \frac{y^n \left|\xi_i\right|^2}{n!}\right]\\
&\underset{}{=} 1 + y\mathbb E\left[\xi_i\right] + \frac12y^2\mathbb E\left[\xi_i^2\right] + y^3\mathbb E\left[\xi_i^2\right]\sum_{n \ge 0} \frac{y^n}{(n+3)!}\\
\end{align}
By simple induction $(n+3)! \ge 6\cdot 3^n$ then 
\begin{align}
\mathbb{E}\left[e^{y\xi_i}\right] &\le 1 + y\mathbb E\left[\xi_i\right] + \frac12y^2\mathbb E\left[\xi_i^2\right] + \frac{1}6y^3\mathbb E\left[\xi_i^2\right]\sum_{n \ge 0} \frac{y^n}{3^n}\\
&= 1 + y\mathbb E\left[\xi_i\right] + \frac12y^2\mathbb E\left[\xi_i^2\right] + \frac{1}6y^3\mathbb E\left[\xi_i^2\right]\frac{1}{1-\frac{y}{3}}
\end{align}
