If $M \subset \mathbb{R}^{n+1}$ is a compact hypersurface, then $M$ is not a minimal hypersurface

Question

I would like to know if there are any conditions on a compact hypersurface $M^n \subset \mathbb{R}^{n+1}$ that ensure that $M$ is not a minimal hypersurface. The motivation for this question is because I would like to know if there is a generalization of this result and a possible generalization of this result is used on page $786$ of this paper, but the hypotheses for this generalization are not clear for me.

Thanks in advance!

$\textbf{EDIT:}$

Following the hint given by Ted, I tried a proof by finite induction on the dimension of $M^n$:

1. Base step ($n = 2$):

Is was done in the linked topic.

2. Induction hypothesis:

a compact hypersurface $M^n \subset \mathbb{R}^{n+1} \Longrightarrow M^n$ is not a minimal hypersurface.

3. Induction step:

I will prove that induction hypothesis remains for $n+1$ arguing by contradiction. Suppose that exists $p \in M$ such that $H(p) = 0$, then

$$\kappa_1(p) + \cdots + \kappa_{n+1}(p) = 0 \Longrightarrow \kappa_{n+1}(p) = -(\kappa_1(p) + \cdots + \kappa_n(p)) \ (\star)$$

Let be $P_{n+1}$ a hyperplane going through $p$ such that the curvature $\kappa_{n+1}(p)$ of $N^n := M^{n+1} \cap P_{n+1}$ is zero, then I can apply the induction hypothesis to $N^n$ to get

$$\kappa_1(p) + \cdots + \kappa_n(p) \neq 0.$$

This and $(\star)$ imply that $\kappa_{n+1}(p) \neq 0$ on $N^n$, which is a contradiction with the fact that $\kappa_{n+1}(p) = 0$ on $N^n$. $\square$

Answer 1

No compact hypersurface in Euclidean space can be minimal. There is always a point of positive Gaussian curvature, and the mean curvature at such a point cannot be $0$ (why?).