On the Project Nayuki article on Barrett reduction, one important lemma is the inequality

$$$\displaystyle\frac{x}{n} - 1 < \frac{xr}{4^k} ≤ \frac{x}{n} \implies \frac{x}{n} - 2 < \left\lfloor \frac{x}{n} - 1 \right\rfloor ≤ \left\lfloor \frac{xr}{4^k} \right\rfloor ≤ \frac{x}{n}$$$

We shall show that this inequality holds even if $$$\dfrac{xr}{4^k}$$$ is replaced by $$$\dfrac{(x-\delta) r}{4^k}$$$, with $$$0 ≤ \delta < 2^{62}$$$.

First we need to obtain a tighter bound on $$$r$$$. Given that $$$r$$$, $$$k$$$, and $$$n$$$ are fixed in our algorithm, we can verify that $$$\dfrac {4^k} n - r < 2^{-9}$$$. Let this value be $$$\varepsilon$$$. We thus obtain $$$\displaystyle \frac{4^k}{n} - \varepsilon < r < \frac{4^k}{n}$$$

Multiply by $$$x-\delta \geq 0$$$: $$$\displaystyle (x-\delta)(\frac{4^k}{n} - \varepsilon) < (x-\delta)r < (x-\delta)\frac{4^k}{n}$$$

Given that $$$\delta$$$ is nonnegative we can relax the right bound to $$$x\dfrac{4^k}{n}$$$ and forget about it thereafter.