I want to prove the following lemma
Lemma (version 1) :
If $a,b \in \mathbb{R}^n$ are such that $|a-b| > |a| > 0$ $|a - b| > |b| > 0$ then the angle between $a,b$ is bounded by below by a costant depending only on $n$, in particolar one has $\big| \frac{a}{|a|} - \frac{b}{|b|} \big| \geq 1$
I tried in many ways but I can't find a way to prove this.
The lemma has been stated in this lecture
Actually I would like to modify the statement, and to prove an inequality of the following type
Lemma (version 2) For each $n$ there exists a $\delta>0$ sufficiently small such that
If $a,b \in \mathbb{R}^n$, are such that $|a - b| > (1 - \delta)|a| > 0$ , $|a - b| > (1 - \delta)|b| > 0$, then the angle between $a,b$ is bounded below by a positive fixed costant depending only on $n$ (and possibly on $\delta$), in particular one has $\big| \frac{a}{|a|} - \frac{b}{|b|} \big| \geq c_\delta$. Where $c_\delta > 0$ depends only on $n$ and on $\delta$
But I think that once one proves version 1 then version 2 is an easy consequence
Since only two vectors $\vec {OA}$ and $\vec {OB}$ are involved, we can diagram $O,A,B$ as coplanar points in the 2D plane, and then use elementary facts about plane Euclidean geometry.
Fix $\alpha =\angle AOB$ and $AB$.
Inscribed Angle Lemma. For a fixed segment $AB$ the locus of all points $C$ for which the angle $\angle ACB =\alpha$ is a pair of congruent circular arcs from $A$ to $B$. This locus is symmetric across the line L that is the perpendicular bisector of $AB$.
(Proof in appendix).
Proof of lecture's first covering lemma.
As you vary $C\in L$, you see that the hypotheses $AC<AB$ forces $C$ to lie inside an equilateral triangle whose sides have length $AB$. And the smallest possible angle that can be subtended in such cases is $\alpha= \pi/3$. This gives the lower bound for angle stated in the lecture you cited.
Appendix. Complex-analytic proof of the inscribed angle lemma. The locus can be regarded as the set of all points $z$ in the complex plane for which $\alpha = |Arg (\frac{(a-z)}{(b-z)})|$. (This follows from the fact that if we use polar notation to write $ a-z= r_1 e^{i \theta _1}$ and $b-z= r_2 e^{i \theta_2}$ then their quotient is a real multiple of $e^{i (\theta_1-\theta_2)}$ and the argument of this expression is $\Delta \theta=\alpha$.)
The linear-fractional map $w= \frac{a-z}{b-z}$ maps circles and lines to circles or lines; the same is true of the inverse map. However, the locus associated to $\alpha$ is obviously not a line! The two rays $Arg w= \pm \alpha$ must correspond to subsets of two circles. After a translation, rotation, and dilation in the $z$ plane we can reduce to the special case in which $a$ and $b$ are the conjugate symmetric pair $a,b=e^{\pm I \alpha/2}$; then confirm that in this case the locus is a pair of circular arcs symmetric under reflection across both coordinate axes. Do this by finding the inverse map and computing the image of the ray $w= t e^{i \alpha}$ in the $z$ plane.)
P.S. There are of course other ways to prove the
Inscribed Angle Theorem
but many have the logical shortcoming of merely showing that the locus is some unspecified subset of the union of two congruent circles. The subtlety is that one should determine exactly what that subset is.