Let $(B_t)_{t\ge 0}$ be a 1-dimensional Brownian motion. We need to show that for any $a,b \gt 0$ we have $$\mathbb P(E) \le e^{-ab}\qquad E:= \{B_t \ge b+ \tfrac12at, \text{for some t} \ge 0\}$$
I have proceeded as follows: first I defined $$M_t=e^{aB_t -a^2 t/2}, t\ge 0$$ which is a martingale. Then $ E=\{M_t\ge e^{ab}, \text{for some }t\ge 0\}$. Then I defined a stopping time $T:=\inf\{t\ge 0:M_t=e^{ab}\}$. Since paths of a Brownian motion are continuous, so are the paths of $M_t$. Furthermore, since every path of BM starts from zero, every path of $M$ starts from $1$ which is $\le e^{ab}$, so $E=\{T<\infty\}$. Now I tried to compute expectations of the stopped processes in order to get the probability that $T$ is finite, since we know by optional stopping they're equal to the expectation of $M_0$ which is one, but couldn't get anything.
Is this approach in the right direction? How to proceed?
$ E=\{M_t\ge e^{ab}, \text{for some t}\ge 0\}$. You are correct about it, as well as the introduction of the stopping time.
Now you can introduce the stopped martingale $N_t=M_{min(\tau,t)}$ . We have $E(N_t)=1$, and $E(N_t)=E(N_t1_{\tau<\infty})+E(N_t1_{\tau=\infty})$, Because $N_t$ is bounded , we can include the limit of t to the infinity term inside the expectation , and because of the continuity of N , we have a.s $N_{\tau}=e^{ab}$.
we have $1=e^{ab}E(1_{\tau<\infty})+E(\lim_{t\rightarrow \infty}N_{t}1_{\tau=\infty})$. We obviously have a.s $N_{t}>0$, then we can conclude