This question has already been asked: Curve on a sphere
but is slightly different because I don´t have the hypothesis that $\alpha(t)$ is a unit-speed curve; anyway I wanted to do it by myself
Prove that a curve $\alpha:I\subset \mathbb R \to \mathbb R^3$ whose image is on the sphere of radius $R$ has a curvature $K\ge 1/R$
My attempt:
$\alpha(t)$ verifies $x^2(t)+y^2(t)+z^2(t)=R^2$ so if we derive this twice: $$x(t)\ddot x(t) +y(t)\ddot y(t)+z(t)\ddot z(t)+ \dot x^2(t)+\dot y^2(t) + \dot z^2(t)=0$$
but we can express the left side as:
$$<\alpha(t),\ddot \alpha(t)>+ <\dot \alpha(t),\dot \alpha(t)>=0$$ which is equal to: $$<\alpha(t),\ddot \alpha(t)>=-<\dot \alpha(t),\dot \alpha(t)>$$ then $$|<\alpha(t),\ddot \alpha(t)>|=|-<\dot \alpha(t),\dot \alpha(t)>|=|<\dot \alpha(t), \dot \alpha(t)>|=||\dot \alpha(t)||^2$$
applaying Cauchy-Schwarz inequality we have:
$$||\alpha(t)|| \cdot ||\ddot \alpha(t)|| \ge |<\alpha(t),\ddot \alpha(t)>|=||\dot \alpha(t)||^2$$ hence : $${{||\alpha(t)|| \cdot ||\ddot \alpha(t)||}\over {||\dot \alpha(t)||^2} }\ge 1$$ $${{||\ddot \alpha(t)||}\over {||\dot \alpha(t)||^2} }\ge {1\over {||\alpha(t)||}}$$
Multiplying and dividing by $||\dot \alpha(t)||$ on the left side of the inequality:
$${{||\dot \alpha(t)|| \cdot ||\ddot \alpha(t)||}\over {||\dot \alpha(t)||^3} }\ge {1\over {R}}$$ because $R=||\alpha(t)||$
I think I´m almost done, I want to have in the left side of the inequality the expression :$${||\dot \alpha(t) x \ddot \alpha(t)||}\over {||\dot \alpha(t)||^3}$$ which is the curvature $K$ of $\alpha(t)$ maybe I can use that fact that $$||\dot \alpha(t) x \ddot \alpha(t)||=||\dot \alpha(t)||\cdot ||\ddot \alpha(t)|| sin\theta$$ but I´m not sure
I could really appreciate if you can help me in the last step :)
