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Source. Lemma 3, arXiv:2406.07218v3, PDF p. 2; proof in Section 2, pp. 3--4.

Used in. Lemma 4, and through it Theorem 1.

Statement

Lemma 3 (p. 2): "For every integer $i\geqslant1000$, at least 1‰ of the numbers in the interval $(1/i,1/(i-1)]$ have non-greedy best two-term Egyptian underapproximations."

In other words: for every integer $i\ge1000$, the set of $x\in(1/i,1/(i-1)]$ whose best two-term Egyptian underapproximation is strictly larger than its greedy one has Lebesgue measure at least $\frac1{1000}\cdot\frac1{(i-1)i}$, one thousandth of the interval's length. The terms are defined on the Theorem 1 page; for $x$ in this interval the greedy two-term underapproximation is $1/i+1/j$ for some $j\ge(i-1)i+1$ (p. 2). The share $1/1000$ is not claimed to be sharp; the paper remarks that only a positive share independent of $i$ is needed.

Proof sketch (pp. 3--4)

A sketch written here. The non-greedy candidates are the sums $1/(i+1)+1/m$ with $m$ between $i(i+1)/2$ and about $3i(i+1)/5$; each such sum, rewritten as $1/i+1/x$ for a real $x$, sits inside one of the gaps between consecutive greedy two-term values, and every point of that gap to its right has a best two-term underapproximation beating the greedy one. A difference argument on consecutive candidates shows that order $i^2$ of them lie a fixed fraction away from the right end of their gap, and the resulting gaps, each of length of order $i^{-4}$, add up to more than one thousandth of the interval. The proof was read for structure only and has not been independently reviewed.

Bears on. #206: a step in the proof of Theorem 1, the result that answers the problem; the lemma alone concerns only two-term underapproximations.