Let $f:\mathbb{R}^n \to \mathbb{R}$ be a smooth convex function. Then for any constant $c>0$, $f + c||x||^2$ is strongly convex. How can I show that the set $\{x \in \mathbb{R}^n : f(x) + c||x||^2 \leq 0\}$, the lower level set of $0$ is uniformly convex. By uniformly convex, I mean that the boundary has principal curvatures have a positive uniform lower bound.
I am trying to understand the approximation of a convex domain by increasing uniformly convex smooth domains, as posed in another question.
Fix $x_0$. Since $f$ is convex, its graph lies above the plane $z=f(x_0)+(x-x_0)\nabla f(x_0)$ to which it is tangent. Therefore, the graph of $g(x)=f(x)+\|x\|^2$ lies above the paraboloid $z=f(x_0)+(x-x_0)\nabla f(x_0)+\|x\|^2$ to which it is tangent. Hence, the curvature of any normal section of the graph of $g$ is no less than the corresponding curvature of the paraboloid.
It remains to check that the paraboloid has positive curvature everywhere. One way, direct but possibly tedious, is to exhibit a sphere tangent to it and containing a piece of paraboloid in its interior. Another way is to argue that the extremal curvatures of a paraboloid, as a surface of revolution, are attained by sections that are either along the axis of rotation or perpendicular to it. See The principal curvatures of a surface of revolution.