Show that $E\exp(-tX_i) \leq \frac{1}{t}$ This is exercise 2.2.10 present in the book High-Dimensional Probability, by Vershynin.
Let $X_1,\ldots,X_n$ be non-negative independent r.v with the densities bounded by $1.$ Show that the MGF of $X_i$ satisfies
$$
E \exp(-tX_i)\leq \frac{1}{t}
$$
After that, deduce that for any $\varepsilon >0$, one has
$$
P\left(
\sum^n_{i=1}X_i \leq \varepsilon n
\right)\leq 
(e\varepsilon)^n
$$
Some help would be much appreciated. I was not able to prove event the first inequality. This question is present in the section dealing with Hoeffding's inequality, so it probably is used somehow.
 A: The first inequality comes from the fact that they have densities bounded by 1:
$$ \mathbb{E}[e^{-t X_i}] = \int_0^\infty e^{-tx} p_i(x)dx \le \int_0^\infty e^{-tx}dx = \frac 1t. $$
For the second inequality, we can show the bound assuming the $X_i$ are independent.  We have that for any $t > 0$
\begin{align*}
P\left( \sum_{i=1}^n X_i \le \varepsilon n \right) &= P\left( \sum_{i=1}^n(-t X_i) \ge -\varepsilon nt \right) \\
&= P\left( \exp \left(\sum_{i=1}^n(-t X_i)\right) \ge e^{-\varepsilon nt} \right) \\
&\le e^{\varepsilon nt} \mathbb{E}\left[\exp \left(\sum_{i=1}^n(-t X_i)\right)\right] \\
&= e^{\varepsilon nt}\prod_{i=1}^n \mathbb{E}[e^{-tX_t}] \\
&\le e^{\varepsilon n t} \frac{1}{t^n} \\
&= \left( \frac{e^{\varepsilon t}}{t} \right)^n.
\end{align*}
Now choose $t$ to minimize $\frac{e^{\varepsilon t}}{t}$.  The value that minimizes $\frac{e^{\varepsilon t}}{t}$ is $t^* := \frac 1{\varepsilon}$, so we have
$$ P\left( \sum_{i=1}^n X_i \le \varepsilon n \right) \le \left( \frac{e^{\varepsilon t^*}}{t^*} \right)^n = (e \varepsilon)^n.$$
