Suppose that $O$ is a bounded open set in $\mathbb{R}^d$ and let $t_1, \ldots, t_n$ denote some distinct points within $O$. I have come across the inequality $$ vol(O) \geq n \min_{i \neq j} ||t_i-t_j||^d V_d, $$ where $V_d$ is the volume of the unit ball in $\mathbb{R}^d$. My question is, is this inequality even true?
It would seem to me that the strategy of proving it, if true, would be to put a ball of radius $\min_{i \neq j} ||t_i-t_j||$ on each one of the $t_i$ and then sum the volumes. But then, there would certainly be overlaps among these balls and some of these volumes may even be outside $O$ so that it's not clear to me if adding all of these volumes would indeed produce this inequality.
Perhaps you need some extra assumptions on $n$ and on the points that you choose. Because for $\mathbb{R}^2$ it does not work even for two points. Take for example $O = (0,1) \times (- \varepsilon, \varepsilon)$ which is open in $\mathbb{R}^2$ and clearly bounded. Then $Vol(O) = 2 \varepsilon $. Consider the points $t_{1}=(\frac{1}{3},0) , t_{2} = (\frac{2}{3},0)$ which are clearly in $O$ then $\|t_{1}-t_{2}\|^2 \cdot V_{2} = \frac{\pi}{9}$ and thus your inequality would imply that:
$$ \varepsilon \geq \frac{2\pi}{9}$$
And you just have to take $\varepsilon = \frac{1}{6}$ and you reach a contradiction.