<s>[INST] Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$ Prove the following for all real $x$
i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$
ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$
For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant, simpler method.
For ($i$) I tried cases like in part ($ii$) but since there are 2 variables that would lead to 4 cases.
An elegant and easy solution will be much appreciated
 [/INST] For (i), define $\{x\} = x - \lfloor x \rfloor$. Then
\begin{align*}
\lfloor 2x \rfloor + \lfloor 2y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor +  \lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor
\end{align*}
and
\begin{align*}
\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor + \lfloor \{x\} + \{y\} \rfloor
\end{align*}
So the inequality to be proven is equivalent to
\begin{align*}
\lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \ge \lfloor \{x\} + \{y\} \rfloor
\end{align*}
But this is true, since
\begin{align*}
\lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \ge \lfloor 2\max(\{x\}, \{y\})\rfloor \ge \lfloor \{x\} + \{y\} \rfloor
\end{align*}
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