Bounds for an Egyptian Fraction Inequality

Question

Question: If $A\geq B>0$ are rational and $x_{1}\leq x_{2}\leq \cdots \leq x_{n}$ are integers such that $A\geq \sum_{j=1}^{n}\frac{1}{x_{j}}\geq B$, then what is an upper bound on $x_{j}$ in terms of $x_{1},x_{2},\ldots, x_{j-1},A,B,$ and $n$? Is the bound sharp?

If $A=B$, then such a bound is provided in E. Landau Uber die Klassenzahl der binaren quadratischen Formen von negativer Discriminante, Math. Ann. 50 (1903), 671--676; doi: 10.1007/BF01444311, eudml. Specifically one has:

Theorem: If $x_{1}\leq x_{2}\leq\cdots\leq x_{n}$ are integers such that $\sum_{j=1}^{n}\frac{1}{x_{j}}=R$, then $x_{j}\leq (n-i+1)/R_{j-1}$ where $R_{0}=R$, and $R_{k}=R_{k-1}-\frac{1}{x_{k}}$.

Answer 1

UPDATE. As pointed out in the comments, an upper bound exists for $x_i$ only if $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$. If we have $B\leq \sum_{j=1}^{i-1} \frac{1}{x_j}<A$, then any sufficiently large $x_i,x_{i+1},\dots,x_n$ will work fine.

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First, notice that the theorem is rather trivial. Indeed, for any $i=1,2,\dots,n$, we have: $$R = \sum_{j=1}^n \frac{1}{x_j} \leq \sum_{j=1}^{i-1} \frac{1}{x_j} + \sum_{j=i}^{n} \frac{1}{x_i} = \sum_{j=1}^{i-1} \frac{1}{x_j} + (n-i+1)\frac{1}{x_i},$$ implying that $$x_i \leq \frac{n-i+1}{R-\sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

Replacing "$R = \sum_{j=1}^n \frac{1}{x_j}$" with "$B \leq \sum_{j=1}^n \frac{1}{x_j}$", we get a similar upper bound for $x_i$ (assuming that $\sum_{j=1}^{i-1} \frac{1}{x_j} < B$): $$x_i \leq \frac{n-i+1}{B-\sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

The bound is tight: it is attained when $\sum_{j=1}^n \frac{1}{x_j} = B$ and $x_i=x_{i+1}=\dots=x_n$.

The value of $A$ can be used to obtain a lower bound for $x_i$. Namely, from $$A\geq \sum_{j=1}^n \frac{1}{x_j} \geq \sum_{j=1}^i \frac{1}{x_j}$$ we get $$x_i \geq \frac{1}{A - \sum_{j=1}^{i-1} \frac{1}{x_j}}.$$

P.S. There is a related Kellogg problem, which for the case $R=1$ bounds $x_n$ uniformly as $x_n \leq s_{n-1}-1$ (where $s_j$ is Sylvester's sequence). This bound can be also extended to lower terms, giving $x_i \leq (n+1-i)\cdot (s_{i-1}-1)$ for all $i=1,2,\dots,n$.