In my studies of differential geometry I am trying to show no compact minimal surfaces exists. This is the last step I am stuck with: Consider regular surfaces $S_1, S_2$ defined by graphs of functions and assume $$f_1(x,y) \leq f_2(x,y)$$ in a neighberhood of the origin, $f_1(0,0)=f_2(0,0)=0$ and that both their tangent planes are the $XY$ plane, that is $T_{(0,0)}S_1=T_{(0,0)}S_2=\{z=0\}$. Prove that at the origin the following inequality is satisfied for the mean curvature: $$H_1(0,0) \leq H_2(0,0)$$
I want to differentiate the Gauss map $N$ and somehow to show that its eigenvalues at $(0,0)$ are larger and hence the mean curvature. I need help with formulating this idea. I will be glad to get some hints and tips.
This is certainly the most difficult way I can imagine to prove that there is no compact minimal surface in $\Bbb R^3$. But the answer to your question is to write out the second-degree Taylor polynomials at $(0,0)$ for $f_1$ and $f_2$. It might help to choose the axes in the directions of the principal directions for one of the surfaces.