A probabilistic angle inequality Conjecture: There is a universal constant $c$ such that for any fixed nonzero real vector $q$ of any dimension $n$ and any random vector $p$ of the same dimension $n$ with independent components uniformly distributed in $[-1,1]$, we have 
$$(p^Tp)(q^Tq)\le cn(p^Tq)^2$$ with probability $\ge 1/2$.
Simulation suggests that the best constant is approximately $c\approx 16/7$.
I'd be interested in techniques for proving this and related statements. In particular, can one determine the best constant $c$?
An application of the inequality to numerical optimization is reported in the paper
M. Kimiaei and Arnold Neumaier,
Efficient global unconstrained black box optimization,
http://www.optimization-online.org/DB_HTML/2018/08/6783.html
 A: Normalize $q$ such that $q^Tq=1$ and $q_i\geq 0$, for all $i=1,\ldots,n$. Let $X_i=q_ip_i$, $i=1,\ldots,n$. We must find an absolute constant $c>0$ such that 
$$P\left(c\left(\sum_i X_i\right)^2\geq \frac{\sum_ip_i^2}{n}\right)\geq \frac 1 2.$$
By the weak law of large number $\frac{\sum_ip_i^2}{n} \to E[p_1^2]<1$ in probability. Therefore, it is sufficient to prove that
$$P\left(\left\vert\sum_i X_i\right\vert< c^{-\frac 1 2} \right)< \frac 1 3,$$
for large enough constant $c>0$.
Assume wlog that $q_1=\max_i q_i$. Since $X_1$ is uniformly distributed in an interval of length $2q_1$ (independently of $\sum_{i>1}X_i$),
$$P\left(\left\vert\sum_i X_i\right\vert< \frac {q_1} 4 \middle \vert \sum_{i>1}X_i\right)\leq \frac 1 4.
$$
This concludes the proof in the case that $q_1$ is bounded away from 0. If $q_1$ is arbitrarily close to 0, then the distribution of $\sum_i X_i$ is arbitrarily close to the standard normal distribution, by Berry-Essen Inequality; therefore choosing $c=100$, say, will do. 
A: $\newcommand{\de}{\delta}
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Note that $\E(p^Tq)^2=q^Tq/3$ and, by the Rosenthal inequality, $\E(p^Tq)^4\ll(q^Tq)^2$. So, by the Paley--Zygmund inequality,
\begin{equation}
 \PP((p^Tq)^2\ge q^Tq/6)\gg\frac{(\E(p^Tq)^2)^2}{\E(p^Tq)^4}\ge a,
\end{equation}
where $a>0$ is a universal constant. Since $p^Tp\le n$, we have 
$$(p^Tp)(q^Tq)\le 6n(p^Tq)^2$$ with probability $\ge a$. ($a$ may be less than $1/2$, though.)
