Let $\alpha : I\rightarrow\mathbb{R}^{n}$ be a curve such that $\alpha (t)\neq\underline{0}$, $\forall t\in I$. Let $t_{0}$ a point such that $\alpha (t_{0})$ is as near as possible to $\underline{0}$. Suppose that $\alpha '(t_{0})\neq\underline{0}$. Prove that $\alpha (t_{0})$ is perpendicular to $\alpha ' (t_{0})$.
I though that, because of the hypotesis, $<\alpha(t_{0}),\alpha(t_{0})> = c\in\mathbb{R}-\lbrace0\rbrace$. So
$$\frac{d}{dt} <\alpha(t_{0}),\alpha ' (t_{0})> = <\alpha '(t_{0}),\alpha(t_{0})>+<\alpha (t_{0}), \alpha '(t_{0})>=2<\alpha(t_{0}),\alpha '(t_{0})> = 0$$
So, because $\alpha '(t_{0})\neq\underline{0}$ we have the assertion.
I have some doubts because I didn't use the fact that $\alpha(t_{0})$ is the most near to the origin. Can someome explain to me if I did something wrong? Thanks before.
You are on the right track!
Define the function $|| \alpha(t) || : I \to \mathbb{R}$. Since $\alpha(t_0)$ is closest point to the origin, it means that $||\alpha(t)||$ attains its minimum at $t_0$. That implies that its derivative at $t = t_0$ is zero, i.e. $$ \frac{d}{dt} ||\alpha(t) ||^2 \; \Big|_{t = t_0} = 2 \cdot \langle \alpha(t_0), \alpha'(t_0) \rangle = 0$$ which means that $\alpha(t_0)$ and $\alpha'(t_0)$ are perpendicular.
Aside: To get the left and right brackets for inner product, use
\langleand\rangle- it makes your stuff look way better.