Surface described by the graph and inequality of forms Suppose that $S,T$ are two surfaces locally described by $(u,v,f_{1}(u,v))$ and $(u,v,f_{2}(u,v))$ where $f_{i}$ are maps from the tangent spaces of $S$ to $\mathbb{R}$ (respectively from T). (here $f_{i}$ are the graphs of the surfaces $S$ and $T$).
Assume that $f_{1}(u,v) - f_{2}(u,v) \geq 0$ for every $(u,v)$ in a neighborhoof of $(0,0)$. Let $F_{1}$ and $F_{2}$ be the second fundamental forms of $S$ and $T$ respectively. Why $F_{1}(w) \geq F_{2} (w)$ for every tangent vector $w$?
I guess this has to do with theory from Hessians but don't see it. (See Why can we think of the second fundamental form as a Hessian matrix?)
 A: I think that's what you mean: Suppose $f_1$ and $f_2$ attain their minimum at $(u,v)=(0,0)$, which implies that $\frac{\partial f_i}{\partial u}(0,0)=\frac{\partial f_i}{\partial v}(0,0)=0$ for $i=1,2$.
Now the surface $S$ is parametrized by $\phi(u,v)=(u,v,f_{1}(u,v))$, and the surface $T$ is parametrized by $\psi(u,v)=(u,v,f_{2}(u,v))$. For the surface $S$, we have
 $\phi_u(0,0)=(1,0,\frac{\partial f_1}{\partial u}(0,0))=(1,0,0)$ and  $\phi_v(0,0)=(0,1,\frac{\partial f_1}{\partial v}(0,0))=(0,1,0)$, which implies that the normal unit vector $n$ at $(0,0)$ is given by
$n=(0,0,1).$
Also, $\phi_{uu}=(0,0,\frac{\partial^2 f_1}{\partial u^2})$, $\phi_{vv}=(0,0,\frac{\partial^2 f_1}{\partial v^2})$, and $\phi_{uv}=(0,0,\frac{\partial^2 f_1}{\partial v \partial u})$. This implies that the coefficient of the second fundamental form is given by 
$$e=\phi_{uu}\cdot n=\frac{\partial^2 f_1}{\partial u^2}, f=\phi_{uv}\cdot n=\frac{\partial^2 f_1}{\partial v \partial u}, g=\phi_{vv}\cdot n=\frac{\partial^2 f_1}{\partial v^2}.$$ That is to say, the matrix representing the second fundamental form is the Hessian of $f_1$. 
With the notation above, take $S$ to be given by $(u,v,u^2+v^2+1)$, and $T$ to be $(u,v,2u^2+2v^2)$. That is, $f_1(u,v)=u^2+v^2+1$ and $f_2(u,v)=2u^2+2v^2$. Note that $f_1(u,v)=u^2+v^2+1\geq 2u^2+2v^2=f_2(u,v)$ locally near $(0,0)$, and both of them has minimum at $(0,0)$. Then by above calculation, the second fundamental form of $S$ is the Hessian of $f_1$, which is given by 
$$\frac{\partial^2f_1}{\partial u^2}\frac{\partial^2f_1}{\partial v^2}-(\frac{\partial^2f_1}{\partial u\partial v})^2=2\cdot 2=4.$$
On the other hand, the second fundamental form of $T$ is the Hessian of $f_2$, which is given by 
$$\frac{\partial^2f_2}{\partial u^2}\frac{\partial^2f_2}{\partial v^2}-(\frac{\partial^2f_2}{\partial u\partial v})^2=(4)(4)=16.$$
