Let $X=(X_t)_{t\geq0}$ be a geometric Brownian motion with drift $\mu$ and volatility $\sigma>0$, and let $Y=(Y_t)_{t\geq0}$ be the associated maximum process of $X$. Under certain conditions, we prove a sharp maximal inequality for the geometric Brownian motion. The method of proof is essentially based on explicit forms of the following optimal stopping problem: Find a stopping time $\tau^*$, if it exists, such that \begin{eqnarray*} \Phi(x,y):=\sup_{\tau}\mathbf{E}^{x,y}\left[Y_\tau-c\int_0^\tau X^\theta_sds\right],\quad c,\theta>0 \end{eqnarray*} where the supremum is taken over all stopping times $\tau$ for the process $X$. The present result complements and extends a similar result proved by Graversen and Peskir [1].