my question is as follows. It is known that a closed smooth curve in $R^2$ is convex iff its (signed) curvature has a constant sign. I wonder if one can characterize smooth convex cones in $R^3$ in a similar way.

Here is the precise statement. I say that an open set $V$ is a cone if $tV \subset V$ for any $t>0$. I say that a cone is smooth if its boundary $\partial V$ is non-empty and, say, $C^\infty$ (except at the origin). So I can define the mean curvature $H$ on the smooth part of the boundary of $V$. Do we have the equivalence between (1) $V$ is convex, and (2) $H$ has a constant sign on $\partial V$?

My question is as follows:

It is known that a closed smooth curve in $\mathbb{R}^2$ is convex iff its (signed) curvature has a constant sign. I wonder if one can characterize smooth convex cones in $\mathbb{R}^3$ in a similar way.

Here is the precise statement. I say that an open set $V$ is a cone if $tV \subset V$ for any $t>0$. I say that a cone is smooth if its boundary $\partial V$ is non-empty and, say, $C^\infty$ (except at the origin). So I can define the mean curvature $H$ on the smooth part of the boundary of $V$.

Do we have the equivalence between: (1) $V$ is convex, and (2) $H$ has a constant sign on $\partial V$?