Curvature Relationship with Norm of the Curve at the Point of Maximum Norm

Continuing on my series of questions, for those following, is the following question:

Let $\alpha: I \to \mathbb{R}^3$ be a regular curve. Suppose that for some $t \in I$ the distance from the origin, $|\alpha(t)|$, attains maximum. Prove that $|\kappa(t)| \geq \frac1{|\alpha(t)|}$.

Intuitively this seems a straightforward and insightful property, a curve where every point has a maximum norm would be on a sphere, and the least curvy would be a great circle, i.e. a circle in the plane, thus the least curvature would be the curvature of a great circle, $\frac1{R} = \frac1{|\alpha(t)|}$. I am not sure of how to show this.

Let $\alpha\colon I\subset\mathbb{R}\longrightarrow \mathbb{R}^3$ be a regular space curve parametrized by arc length with the property that: $$\exists\ t_0\in I\ :\ \mathrm{max}_{t\in I}||\alpha(t)||=||\alpha(t_0)||.\ \ \ (\dagger)$$ For convenience we define the square of the distance function $$f\colon I\subset\mathbb{R}\longrightarrow \mathbb{R}^3,\ \ t\mapsto f(t):=||\alpha(t)||^2.$$ By differentiating we obtain $$f'(t)=2\langle \alpha'(t),\alpha(t)\rangle\ \ \&\ \ f''(t)=k(t)\langle \vec{n}(t),\alpha(t)\rangle+1,$$ where $k$ and $\vec{n}$ is the curvature and the principal normal of the curve $\alpha$ respectively. Therefore due to $(\dagger)$ we obtain that: $$f'(t_0)=0\ \ \&\ \ f''(t_0)\leq 0.\ \ \ (\ddagger)$$ Thus it follows: $$|k(t_0)|\geq \frac{1}{|\langle \vec{n}(t_0),\alpha(t_0)\rangle|}.\ \ \ (\star)$$ But $$\alpha(t_0)=\lambda\ \alpha'(t_0)+\mu\ \vec{n}(t_0)+\nu\ \vec{b}(t_0).$$ (where $\{\alpha',\vec{n},\vec{b}\}$ is the Frenet frame of the curve) Since $f'(t_0)=0$ we have that $\lambda=0$. Therefore $$||\alpha(t_0)||^2=\mu^2+\nu^2\geq \mu^2\Rightarrow ||\alpha(t_0)||\geq|\mu|=|\langle \vec{n}(t_0),\alpha(t_0)\rangle|\Rightarrow \boxed{\frac{1}{|\langle \vec{n}(t_0),\alpha(t_0)\rangle|}\geq \frac{1}{||\alpha(t_0)||}}.$$ Now, the desired result follows directly from $(\star)$.