To elaborate on Marek's (correct) answer, since it seems that math is the issue that @Casebash is having:
Start with an integral representing the time elapsed on the moving observer's clock in terms of the stationary observer's coordinates (I'm supressing G and c below. Feel free to replace all $t$'s with $c\,t$'s, and all $M$'s with $\frac{GM}{c^{2}}$'s):
$$\begin{equation}
\Delta \tau = \int d\tau \sqrt{{\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}}
\end{equation}$$
Where $(t,r,\theta,\phi)$ are all considered to be functions of $\tau$ denoting the moving observer's position at time $\tau$. Our problem amounts to looking for the path beginning at $(t_{0},R,\Theta,\Phi)$ and ending at $(t_{f},R,\Theta,\Phi)$ that maximizes this integral, subject to the constraint that the quantity under the square root will be equal to $1$ when the calculation is complete. To make our calculations easier, note that the square root is a monotonic function over all its domain, so we might as well maximize
$$\begin{equation}
\frac{1}{2}\int d\tau \left({\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}\right)
\end{equation}$$
which is considerably simpler. It would take a whole page to carefully vary (meaning that we basically, but not quite, take the derivative of the function with respect to) this with respect to the four independent functions. So, I'm going to just give you a taste of the task by varying with respect to $\theta$. The path of maximum "moving clock" time (heretofore called 'proper time') will be the one for which these variations are zero. The other four functions follow just as easily. The variation of this integral with respect to $\theta$ gives us
$$\begin{align}
\delta \Delta\tau |{}_{\theta}&=\int d\tau\left(-r^{2}{\dot \theta}\delta {\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\delta \theta\right)\\
0&=\int d \tau \left(-\frac{d}{d\tau}\left({\dot \theta} r^{2}\delta \theta\right) + \left({\ddot \theta}r^{2} + 2 r {\dot r}{\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\right)\delta \theta\right)\\
&=\left(\dot \theta(t_{0})r(t_{0})^{2}\delta \theta (t_{0})-\dot \theta(t_{f})r(t_{f})^{2}\delta \theta (t_{f})\right)\\
&\quad + \int d\tau\; r^{2}\left({\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}\right)\delta \theta
\end{align}$$
The first term, that we obtained by integrating the total derivative, vanishes since the variation of $\theta$ is zero at the endpoints $t_{0}$ and $t_{f}$ (our space of states is paths beginning and ending at a fixed value of $\theta$. If the variation will be zero, then the remaining stuff under the integral must turn out to be zero if it is going to be a minimum for an arbitrary fixed-point variation (if the integrand isn't zero, just imagine making the variation a hundred kajillion only at the point where it is nonzero--clearly this isn't an extremum). Therefore, the $\theta$ variable of the maximum proper time path must satisfy ${\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}=0$. It turns out that this equation is exactly the same equation you get for geodesic motion of a particle, which is the path followed by an observer being acted upon only by gravity. A simple calculation of the first integral for any other path (and the fact that it will come out higher) will then show you that this, in fact, is the maximum time path to travel.
