Let $B$ as a $n\times n$ matrix where
$$B_{ij}(\pmb{\theta})=(\theta_i-\theta_j)\sin(\theta_i-\theta_j), 1\leq i<j\leq n$$ and other entries equals to $0$, and $$\theta=[\theta_1,\cdots,\theta_n]\in\{\theta | |\theta_i-\theta_j|\leq \pi/2, \forall i,j\} \setminus \{ \theta | |\theta_i-\theta_j| = 0 \forall i,j\}.$$
Let $b(\pmb{\theta})$ as the below $\frac{n(n-1)}{2}$-dimensional vector
\begin{equation}\label{b}
[(\theta_1-\theta_2)\sin(\theta_1-\theta_2),
\cdots,
(\theta_1-\theta_n)\sin(\theta_1-\theta_n),\\
(\theta_2-\theta_3)\sin(\theta_2-\theta_3),
\cdots,
(\theta_2-\theta_n)\sin(\theta_2-\theta_n),\\
(\theta_3-\theta_1)\sin(\theta_3-\theta_1),
\cdots,
(\theta_3-\theta_n)\sin(\theta_3-\theta_n),\\
\cdots,
(\theta_{n-1}-\theta_n)\sin(\theta_{n-1}-\theta_n)]^T
\end{equation}
From numerical experiement, we observe that the ratio between the $l_1$ norm and $l_2$ norm grows at least $n^{1.7}$
Consider the example $\theta=[\pi/2,0,\cdots,0]$ given by @Iosif Pinelis, we modify the conjecture to at least $n-1$, or modify the question to How to estimate the asymptotic value of this ratio?.
$$\frac{\|b\|_1^2}{\|b\|_2^2}=\frac{\Big(\sum_{i<j}(\theta_i-\theta_j)\sin(\theta_i-\theta_j)\Big)^2}{\sum_{i<j}(\theta_i-\theta_j)^2\sin^2(\theta_i-\theta_j)}\geq n^{1.7}$$
However I have no idea how to prove it.
I would appreciate any hint/help/comment!
(Numerical experiment: for each $n$, we sample $1000$ number of $\theta$ and compute the ratio $\frac{\|b\|_1^2}{\|b\|_2^2}$ on these $\theta$. Then take the average of the obtained ratios. Thus, in the figure, each blue dot represent the average of ratios on $1000$ samples $\theta$, corresponding to $n$. Here $n$ ranges from $4$ to $500$. The ratio shows a growth at least $n^{1.7}$.)
What I tried:
Take the log to obtain $$$$
Best Answer
This conjecture is not true. E.g., if $\theta_1=\cdots=\theta_{n-1}=0$ and $\theta_n=\pi/2$, then the ratio in question is $n-1$.