Inner product of unit vector with a vector uniformly distributed on a hypersphere Let $v$ be some vector in $\mathbb R ^n$ and $a$ a random vector uniformly distributed on the sphere in $n$ dimensions with a radius of $\sqrt n$. Show that:
$$\mathbb P(|\langle a, v \rangle| \geq 1/2 \|v\|_2) \geq 1/2$$
What I've tried
Since $v$ is arbitrary, by symmetry we can take the first coordinate axis unit vector ($e_1$) instead.
$$\mathbb P(|\langle a, v \rangle| \geq 1/2 \|v\|_2) = \mathbb P(|\langle a, e_1 \rangle| \geq 1/2)$$
Now since $\|a\|_2 = \sqrt n$ then:
$$\mathbb P(|\langle a, e_1 \rangle| \geq 1/2) = \mathbb P\left(|\langle s, e_1 \rangle| \geq \frac{1}{2\sqrt n}\right)$$
where $s$ is uniformly distributed on the surface of the unit sphere. I now need to lower bound this expression. Since $\langle s, e_1 \rangle$ is distributed as $|Z_1| / \|Z\|_2$ for $Z \sim N(0, 1)$ and since $\|Z\|_2$ concentrates around $\sqrt n$ then we can argue that the probability in question is at least $1/2$, but I need a non-asymptotic result (for all $n$, not just large $n$).
My more formal attempt is to use the Paley-Zygmund inequality, but that might be overkill for this application. I'm still struggling to evaluate (or get an upper bound for) the expectation of $X = |Z_1| / \|Z\|_2$. The expectation of $X^2$ is easy by linearity of expectation: $1/n$
 A: We shall follow a geometric approach.
Let $S\subset\Bbb R^n$ be a sphere of radius $r=\sqrt{n}$ centered at the origin $0$ of $\Bbb R^n$. If $v=0$ then $|\langle a, v \rangle|\ge 0$ for any $a\in S^{n-1}$. If $v\ne 0$ then it is easy to check that a set $\{a\in S:|\langle a, v \rangle| \geq 1/2 \|v\|_2\}$ consists of two spherical caps, cut from $S$ by hyperplanes orthogonal to $v$ and placed at a distance $1/2$ from the origin. It follows that each cap is spanned by an angle $2\psi$, where $\cos\psi=\tfrac 1{2r}=\tfrac 1{2\sqrt{n}}$.
When $n=1$ then $S$ consists of two points covered by the caps. When $n=2$ then $S$ is a circle, $\cos\psi=\tfrac 1{2\sqrt{2}}<\tfrac 1{1\sqrt{2}}=\cos\tfrac{\pi}4$, so $\psi>\tfrac{\pi}4$  and the caps cut more than half of the circle length.
So from now we assume $n\ge 3$.
We have that ($n-1$) dimensional area of $S$ equals $$A_n=r^{n-1}\frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}.$$
According to [Li], an area $C_n$ of the spherical cap equals $$r^{n-1}\frac{2\pi^{(n-1)/2}}{\Gamma\left(\tfrac {n-1}2\right)}\int_0^\phi\sin^{n-2} \theta d\theta,$$ where, as I understood, $2\phi$ is the angle spanned by the cap.
Thus we have to show that the area remaining after cutting the caps is at most half  of the sphere area, that is
$\frac{2\pi^{(n-1)/2}}{\Gamma\left(\tfrac {n-1}2\right)}\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \frac 14 \frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}$
$\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}.$
Since $n\ge 3$,
$$\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \int_\psi^{\pi/2}\sin \theta d\theta=-\cos\frac \pi{2}+\cos \psi=\tfrac 1{2r}=\tfrac 1{2\sqrt{n}}.$$
Now we estimate the right-hand side of the inequality.
By Legendre’s formula, $$\Gamma\left(\tfrac {n-1}2\right)\Gamma\left(\tfrac n2\right)=\frac{\sqrt{\pi}}{2^{n-2}}
\Gamma(n-1)=\frac{(n-2)!\sqrt{\pi}}{2^{n-2}}.$$
1)) If $n$ is even then $\Gamma\left(\tfrac n2\right)=\left(\tfrac n2-1\right)!$, so
$$\frac{\Gamma\left(\tfrac {n-1}2\right)}{\Gamma\left(\tfrac n2\right)}= \frac{(n-2)!\sqrt{\pi}}{2^{n-2}\left(\tfrac n2-1\right)!^2}=\frac{\sqrt{\pi}}{2^{n-2}}{n-2\choose \tfrac {n-2}2}.$$
Robbins’ bounds imply that  for any positive integer $m$
$$\frac {4^{m}}{\sqrt{\pi m}}\exp\left(-\frac {1}{8m-1}\right)<{2m\choose m}<\frac {4^{m}}{\sqrt{\pi m}}\exp\left(-\frac {1}{8m+1}\right).$$
So
$$\frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}=\frac{\pi}{2^n}{n-2\choose \tfrac {n-2}2}>
\frac{\pi}{2^n} \frac {2^{n-2}}{\sqrt{\pi (n-2)/2}}\exp\left(-\frac {1}{4n-9}\right)=$$
$$\sqrt{\frac{\pi}{8n-16}}\exp\left(-\frac {1}{4n-9}\right)>\sqrt{ \frac{\pi}{8n}}\exp\left(-\frac {1}{7}\right)
>\frac{0.54}{\sqrt{n}}>\frac 1{2\sqrt{n}}.$$
2)) If $n$ is odd then $\Gamma\left(\tfrac {n-1}2\right)= \left(\tfrac {n-3}2\right)!$, so
$$\frac{\Gamma\left(\tfrac {n-1}2\right)}{\Gamma\left(\tfrac n2\right)}=
\frac{2^{n-2}\left(\tfrac {n-3}2\right)!^2}{(n-2)!\sqrt{\pi}}=\frac{2^{n-2}}{(n-2)\sqrt{\pi}}{n-3\choose \tfrac {n-3}2}^{-1}.$$
If $n=3$ then the latter expression equals $\tfrac {2}{\sqrt\pi}>\tfrac 1{2\sqrt{n}}$. If $n>3$ then Robbins’ bounds imply that
$$\frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}=
\frac{2^{n-4}}{n-2}{n-3\choose \tfrac {n-3}2}^{-1}>\frac{2^{n-4}}{n-2}\cdot \frac {\sqrt{\pi (n-3)/2}}{2^{n-3}}\exp\left(\frac {1}{4n-11}\right)>\frac {\sqrt{\pi (n-3)/2}}{2(n-2)}> \frac 1{2\sqrt{n}}.$$
References
[Li] S. Li, Concise Formulas for the Area and Volume of a Hyperspherical Cap, Asian Journal of Mathematics and Statistics 4:1 (2011) 66–70.
