Can anyone help? My approach was to transform this to vector language for unit vectors on the sphere and apply induction but I failed.

First, note that $\cos (2\arcsin (\frac{\sqrt{6}}{3})) =\frac{-1}{3}$, so we need to show that the cosine of at least one such angle is smaller than $\frac{-1}{3}$. Suppose to the contrary that the cosines of all such angles are greater than $\frac{-1}{3}$. Consider any three points from those $n$ points. If there's a semi-sphere that contain all three points, there's a point from other $n-3$ point that lie on the opponent semi-sphere. If there's no semi-sphere that contain all three points, choose any other points from other $n-3$ points. The vectors from the origin to four points we chose will partition the whole sphere into $4$ disjoint parts. There's a part with volume at least $\frac{1}{4}$ of the sphere. Let the three vectors from the origin that form this part pointed at three points $V_1,V_2,V_3$ on the sphere. Let three spherical angle of spherical triangle $V_1V_2V_3$ are equal to $A,B,C$ degree. It's well known that the volume of the part is equal to $\frac{A+B+C-180}{360}$ times the total volume of the sphere. Hence, $A+B+C\geq 360$. And note that $A,B,C\in (0,180]$. Also, by spherical trigonometry, $\cos (V_iOV_j)$ where $i\neq j\in \{ 1,2,3\}$ is equal to $\frac{\cos (A)+\cos (B) \cos (C)}{\sin (B)\sin (C)}$ or two other cyclic permutation of $A,B,C$. If there exists, WLOG it's $A$, angle $A$ that $\cos (A)>0$, which means $A<90$. We get that $\cos (B)\leq 0$, otherwise $B<90$ will leads to $C>180$, impossible. Similarly, $\cos (C)\leq 0$. In summary, $A\in (0,90)$ and $B,C\in [90,180]$. We know that $\frac{\cos (B)+\cos (A) \cos (C)}{\sin (A)\sin (C)} > \frac{-1}{3}$, which gives $3\cos (B)+2\cos (A) \cos (C)\geq -\cos (C-A)$. So, $-\cos (C-A)\leq 0$, this gives $C-A\leq 90$. Similarly, $C-B\leq 90$. Hence, $360\leq A+B+C\leq 3A+180\Rightarrow A\geq 60$. So, $-cos (C-A)\leq 3\cos (B)+2\cos (60)\cos (C)\leq 0+\cos (C)=\cos (C)$. We get $\cos (C) +\cos (C-A)=2\cos (C-\frac{A}{2})\cos (\frac{A}{2}) \geq 0\Rightarrow C-\frac{A}{2}\leq 90$. Similarly, $B-\frac{A}{2}\leq 90$. Hence, $360\leq A+B+C\leq 2A+180\Rightarrow A\geq 90$, contradiction. Now, we can conclude that $\cos (A),\cos (B),\cos (C)\leq 0$. In other words, $A,B,C\in [90,180]$. Let $p,q,r\in [0,90]$ that $p=180-A,q=180-B,r=180-C$. We've $p+q+r=540-(A+B+C)\leq 180$. And WLOG $p\geq q\geq r$. Transform our inequalities to $\cos (p)\cos (q) -\cos (r) > \frac{-1}{3}\sin (p) \sin (q)$ and two other cyclic permutations. Hence, $\cos (p)\cos (q) +\frac{1}{3}\sin (p) \sin (q)\geq \cos (r)$. By C-S, we get $\cos^2 (r)\leq \Big( \cos^2 (p)+\frac{1}{3}\sin^2 (p)\Big) \Big( \cos^2 (q)+\frac{1}{3}\sin^2 (q)\Big) = \Big( 1-\frac{2}{3}\sin^2 (p)\Big) \Big( 1-\frac{2}{3}\sin^2 (q)\Big)$. Since $p\geq q\geq r$, we easily get $p\geq 60\Rightarrow \sin (p)\geq \frac{\sqrt{3}}{2}$ and $\cos (r)\geq \cos (q)$. Hence, $\cos^2 (q)\leq \frac{1}{2}\times \Big( 1-\frac{2}{3}\sin^2 (q)\Big) \Rightarrow 1\geq 2\cos^2 (q)+\frac{2}{3}\sin^2 (q)\Rightarrow \cos (q)\leq \frac{1}{2}$. This gives $q\geq 60$. So, $\sin (q)\geq \frac{\sqrt{3}}{2}$ too. So, $\cos^2 (r)\leq \frac{1}{4}\Rightarrow \cos (r)\leq \frac{1}{2}$. This gives $r\geq 60$. But we've $p+q+r\leq 180$, so $p=q=r=60$. But that case gives us $\cos (p)\cos (q) -\cos (r) = \frac{-1}{3}\sin (p) \sin (q)$, contradiction since the sign must be strictly greater. This contradiction completes the proof.