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

Let's denote with $\{ x \}$ the fractional part:
$$
x = \left\lfloor x \right\rfloor  + \left\{ x \right\}
$$
Then I would suggest that you first  master the addition
$$
\eqalign{
  & \left\lfloor {x + y} \right\rfloor  = \left\lfloor {\left\lfloor x \right\rfloor  + \left\{ x \right\} + \left\lfloor y \right\rfloor  + \left\{ y \right\}} \right\rfloor  =   \cr 
  &  = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor  =   \cr 
  &  = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \cr} 
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Thereafter you simply have
i)
$$
\left\lfloor {2x} \right\rfloor  = \left\lfloor {x + x} \right\rfloor  = 2\left\lfloor x \right\rfloor  + \left[ {1/2 \le \left\{ x \right\}} \right]
$$
so
$$
\left\{ \matrix{
  \left\lfloor {2x} \right\rfloor  + \left\lfloor {2y} \right\rfloor  = 2\left\lfloor x \right\rfloor  + 2\left\lfloor y \right\rfloor  + \left[ {1/2 \le \left\{ x \right\}} \right] + \left[ {1/2 \le \left\{ y \right\}} \right] \hfill \cr 
  \left\lfloor {x + y} \right\rfloor  + \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  = 2\left\lfloor x \right\rfloor  + 2\left\lfloor y \right\rfloor  + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \cr}  \right.
$$
and clearly
$$
\left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \le \left[ {1/2 \le \left\{ x \right\}} \right] + \left[ {1/2 \le \left\{ y \right\}} \right]
$$
ii)
$$
\eqalign{
  & \left\lfloor x \right\rfloor  = \left\lfloor {x/2 + x/2} \right\rfloor  =   \cr 
  &  = 2\left\lfloor {x/2} \right\rfloor  + \left[ {1/2 \le \left\{ {x/2} \right\}} \right] \cr} 
$$
