shobber wrote: Find all positive integers $m$ and $n$ such that the inequality \[\lfloor (m+n) \alpha \rfloor+ \lfloor (m+n) \beta \rfloor \ge \lfloor m \alpha \rfloor + \lfloor n \beta \rfloor + \lfloor n(\alpha+\beta) \rfloor\] is true for any real numbers $\alpha$ and $\beta$. Here $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$. Are $\alpha$ and $\beta$ really supposed to be real? Not just positive real? Since for real numbers $\alpha$ and $\beta$, the questions becomes quite easy.. Proof. Let $\alpha$, $\beta$ be integers. Then the inequality $(m-n)\beta \ge 0$ must hold true. However, if $m>n$, we can choose a negative $\beta$, if $m<n$, we choose a positive integer $\beta$. Consequently, $m=n$. The inequality becomes \[\lfloor 2m \alpha \rfloor + \lfloor 2m \beta \rfloor \ge \lfloor m \alpha \rfloor + \lfloor m \beta \rfloor + \lfloor m(\alpha+ \beta) \rfloor .\] Define integers $A$ and $B$, such that we have $A > m \alpha \ge A-1$ and $B>m \beta \ge B-1$. Now we can split the problem up into three cases: $\bullet$ $2A > 2m \alpha \ge 2A-1$ and $2B > 2m \beta \ge 2B-1$, $\bullet$ $2A > 2m \alpha \ge 2A-1$ and $2B-1 > 2m \beta \ge 2B-2$, $\bullet$ $2A-1 > 2m \alpha \ge 2A-2$ and $2B-1 > 2m \beta \ge 2B-2$. A short computation shows that the inequality holds true in each of the three cases. Hence every pair $(m$, $n)=(t$, $t)$ for some positive integer $t$ fits. $\square$