Truncation-loss conjecture for quasi-regular distributions

Let $\boldsymbol{F}=\{F_i\}_{i\in[n]}$ be $n\geq1$ asymmetric quasi-regular valuation distributions, let $\varepsilon\in(0,1)$, and define truncated distributions $\widetilde{\boldsymbol{F}}=\{\widetilde F_i\}_{i\in[n]}$ by truncating each distribution to $[0,t]$, where

$$t=\frac{1}{\varepsilon}\,\operatorname{BOM}(\boldsymbol{F}).$$

Here $\operatorname{BOM}(\boldsymbol{F})$ is the expected revenue of the Bayesian Optimal Mechanism. Quasi-regular truncation conjecture. The truncated distributions satisfy

$$\operatorname{BOM}(\widetilde{\boldsymbol{F}})\geq(1-O(\varepsilon))\operatorname{BOM}(\boldsymbol{F}).$$

This would show that truncating quasi-regular distributions incurs the same-order revenue loss as truncating regular distributions, yielding the corresponding sample-complexity bound. The conjecture is proved for single-buyer instances, $n=1$, but remains open for multiple buyers.