I am trying to prove an analogue to Fermat's principle that light rays travel along the path of least time for a static metric. First, I want to show that a null geodesic for a static metric (i.e. all metric components are independent of coordinate time and $g_{\mu 0} = 0 \hspace{0.4cm} \forall \mu \neq 0$) can be written as
$\frac{g_{jk}}{g_{00}} \frac{d^2 x^k}{dt^2} + \frac{1}{2g_{00}} \left(\partial_l g_{jk} + \partial_k g_{jl} - \partial_j g_{kl} \right) \frac{dx^k}{dt} \frac{dx^l}{dt} = 0$
where the latin indices only run through spatial components. It is my understanding that coordinate time is an affine parameter for a static metric when considering null paths. Thus, the criteria for null paths can be written as
$g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt} = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} g_{00} + g_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt} = 0$
Since coordinate time is an affine parameter, we can do some algebra in the geodesic equation to get
$\frac{d^2x^i}{dt^2} + \Gamma^i_{00} + \Gamma^i_{jk} \frac{dx^j}{dt} \frac{dx^k}{dt} = 0\\ \frac{d^2x^i}{dt^2} - \frac{1}{2} g^{il} \partial_l g_{00} + \Gamma^i_{jk} \frac{dx^j}{dt} \frac{dx^k}{dt} = 0\\ \frac{g_{mi}}{g_{00}}\frac{d^2x^i}{dt^2} - \frac{g_{mi}}{g_{00}} \frac{1}{2} g^{il} \partial_l g_{00} + \frac{g_{mi}}{g_{00}}\Gamma^i_{jk} \frac{dx^j}{dt} \frac{dx^k}{dt} = 0 $
From here I am pretty lost. I realize that I essentially need to get rid of the middle term in the geodesic equation, but I'm not sure how the null path criteria helps me do that, and I've tried several methods of substitution. It also does not make intuitive sense to me why the middle term would go to zero. Is there anything wrong with one of the assumptions I've made above, or is this purely an algebraic issue that I cannot work out? Any help is much appreciated.
