How to use Minkowski's formula to prove Liebmann's theorem? Let $M$ be a compact connected surface embedded in $\Bbb R^3$, and has constant Gaussian curvature. I am asked to show that $M$ is a canonical sphere by using the Minkowski's formula.
Minkowski's formulas are: the integral of $H+pK$ on surface equals $0$, and the integral of $pH+1$ on surface equals $0$, where $p=\langle n,x\rangle$ is the support function of $M$, $H$ is the average curvature with respect to $n$ (normal fields on $M$) and $K$ is the Gaussian curvature of $M$. And $x$ is the position function. I want to show that $H^2-K=0$ but can't get that easily.
 A: 
Minkowski's formulas:
  $$\int_{\Sigma}HdA=\int_{\Sigma}KpdA$$
  $$\int_{\Sigma}dA=\int_{\Sigma}HpdA$$
And $p=-<r,n>$

(1)Let us think about the case: $K$ is constant.

First, because a compact surface must have a positive $K$ at some point, then we assume $K>0$. And from Hadamard Theorem, we know that $\Sigma$ is convex. So we can choose a $n$ to let $p>0$ by the property of convex surface.
Second, because 
  $$H^2=(\frac{k_1+k_2}{2})^2\geq{k_1k_2}=K$$
  So $H\geq\sqrt{K}$.
  Then we have 
  $$\int_{\Sigma}KpdA=\int_{\Sigma}HdA\geq\int_{\Sigma}\sqrt{K}dA=\sqrt{K}\int_{\Sigma}dA=\sqrt{K}\int_{\Sigma}HpdA\geq{K}\int_{\Sigma}pdA$$
  So $H^2=K$.

(2)Let us think about the case: $H$ is constant. (We assume $\Sigma$ is convex surface)

$$\int_{\Sigma}KpdA=\int_{\Sigma}HdA=H\int_{\Sigma}dA=H\int_{\Sigma}HpdA=\int_{\Sigma}H^2pdA$$
  $$\int_{\Sigma}(H^2-K)pdA=0$$
Because $H^2-K=\frac{1}{4}(k_1-k_2)^2\geq0$, then $H^2=K$ and it is an umbilical point surface. So it is a sphere.

This comes from what you think.
