Let $(X_{j})_{j \geq 1}$ be a sequence of random variables with $X_{j}$ having mean zero and a finite moment generating function $\phi_{j}(\xi) = E(e^{\xi X_{j}})$ for all $\xi$ in a neighborhood $J$ of zero, $j \geq 1$.
Define $S_{n} = X_{1} + \dots X_{n}$, $Z_{n} = e^{\xi S_{n}}/\prod_{j=1}^{n} \phi_{j}(\xi)$, $M_{n} = \max\{S_{1}, \dots, S_{n}\}$, $m_{n} = \min\{S_{1}, \dots, S_{n}\}$.
Note that $Y_{n} = e^{\xi S_{n}}$ is a submartingale for every $\xi \in J$. Show that for all $\lambda > 0$, $$P(M_{n} \geq \lambda) \leq e^{-\xi \lambda} \prod_{j=1}^{n} \phi_{j}(\xi), \ \xi > 0, \ \xi \in J$$ and $$P(m_{n} \leq -\lambda) \leq e^{\xi \lambda} \prod_{j=1}^{n} \phi_{j}(\xi), \ \xi < 0, \ \xi \in J.$$
I tried to apply Doob's maximal inequality to $Y_{n}$ but I couldn't derive the expressions wanted.
Hint: (For $\xi > 0$) (I am assuming $X_i$ are independent).
1) $$\{M_n \geq \lambda\} \iff \{e^{\xi M_n} \geq e^{\xi\lambda }\}$$
2) $$e^{\xi M_n} = e^{\xi \max_i S_i} = \max_i e^{\xi S_i} = \max_i Y_i$$
Can you take it from here? Let me know if you need additional help.
Update: For the next park, work with $-X_i$. Then $T_n =-S_n = \sum_{i=1}^n -X_i$. And $T_n$ is a martingale. As you mentioned in comments, $$-m_n = \max\{-S_1,-S_2,\cdots, -S_n\}= \max\{T_1,T_2,\cdots, T_n\}$$
You can, using the logic above, show $$P[m_n \leq -\lambda]= P[-m_n \geq \lambda]= P[e^{-m_n\xi} \geq e^{\lambda \xi}]$$ (Here $\xi > 0$). Noting that $e^{T_n\xi}$ is a submartingale. $$P[e^{-m_n\xi} \geq e^{\lambda \xi}] \leq E[e^{T_n\xi}] e^{-\xi\lambda}= E[e^{-S_n\xi}] e^{-\xi\lambda}$$ Now replace $\xi$ with $-\xi$