Given a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Let $\{B_t:t\ge0\}$ be a standard Brownian Motion. We know that: $$\{\limsup B_n/\sqrt{n}\ge c\}\supseteq \{B_n/\sqrt{n}> c\text{ i.o.}\}$$
Could it be shown that: $$\mathbb{P}(\{B_n/\sqrt{n}> c\text{ i.o.}\})\ge \mathbb{P}(\{B_1>c\})$$? If so, how? Any hint?
Let $\{A_n\}$ be a sequence of events. Since for each $n\ge 1$, $\mathsf{P}\!\left(\bigcup_{m\ge n}A_m\right)\ge \sup_{m\ge n}\mathsf{P}(A_m)$, \begin{align} \mathsf{P}(\limsup_{n\to\infty}A_n)&=\mathsf{P}\!\left(\bigcap_{n\ge 1}\bigcup_{m\ge n} A_m\right)=\lim_{n\to\infty}\mathsf{P}\!\left(\bigcup_{m\ge n}A_m\right) \\ &\ge \lim_{n\to\infty}\left(\sup_{m\ge n}\mathsf{P}(A_m)\right)=\limsup_{n\to\infty}\mathsf{P}(A_n), \end{align} where the second equality follows from the continuity from above. Applying this result to your case, one gets $$ \mathsf{P}(B_n/\sqrt{n}>c\text{ i.o.})\ge \limsup_{n\to\infty}\mathsf{P}(B_n/\sqrt{n}>c)=\mathsf{P}(B_1>c) $$ because $B_n/\sqrt{n}\overset{d}{=}B_1$.