I'm having some conceptual misunderstandings of the Michelson-Morley experiment. The time for the beam going perpendicular ($t_{across}$) of the aether wind I am getting:
$$(ct)^{2}=d^{2}+(vt)^{2}\\ (ct)^{2}-(vt)^{2}=d^{2}\\ c^{2}t^{2}-v^{2}t^{2}=d^{2}\\ t^{2}(c^{2}-v^{2})=d^{2}\\ t^{2}=\frac{d^{2}}{(c^{2}-v^{2})}\\ t_{across}=\frac{d}{\sqrt{(c^{2}-v^{2})}}\\ t_{acrossTotal}=\frac{2d}{\sqrt{(c^{2}-v^{2})}}\\ t_{acrossTotal}=\frac{2d\sqrt{(c^{2}-v^{2})}}{c^{2}-v^{2}}$$
For the time of the beam going with the aether wind (t_along) I get: $$t_{1}=\frac{d}{c+v}\\ t_{2}=\frac{d}{c-v}\\ t_{alongTotal}=\frac{d}{c+v}+\frac{d}{c-v}\\ t_{alongTotal}=\frac{d(c-v)}{(c+v)(c-v)}+\frac{d(c+v)}{(c+v)(c-v)}\\ t_{alongTotal}=\frac{dc-dv+dc+dv}{(c+v)(c-v)}\\ t_{alongTotal}=\frac{2dc}{c^{2}+v^{2}}\\$$
Now from what I'm reading about the experiment, what I need to be showing is that $$t_{along} \geq t_{across}$$ The equations I have now doesn't really show that so I'm confused. From what I'm reading they did the experiment multiple times; by rotating the interferometer, and by waiting for the earth to be in a different position around the sun to detect a .6 fringe difference in the interference patten.
Am I assuming correctly that in order to prove this mathematically, c should greater than or equal to $\sqrt{c^2-v^2}$? Is the experiment trying to show that $t_{along} \geq t_{across}$ when comparing if the aether wind speed is $v=0$ or $v!=0$?
Any assistance in clearing up my misunderstanding is appreciated.