Answer by coffeemath for Floor Inequalities
Let $\theta_i=\{x_i\}$, so that the second inequality reads$$\sum_{i=1}^{n}\left \lfloor q_i \theta_i \right \rfloor \geq \left \lfloor \sum_{i=1}^{n}\theta_i \right \rfloor. \tag{1}$$Also let $L$...
View ArticleFloor Inequalities
Proving the integrality of an fractions of factorials can be done through De Polignac formula for the exponent of factorials, reducing the question to an floored inequality.Some of those inequalities...
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