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
 A: Part ii:
In general: $$\lfloor z\rfloor \leq z< \lfloor z\rfloor+1.$$ 
For ii, set $z=x/2$ then we get:
$$\left\lfloor\frac{x}{2}\right\rfloor\leq \frac{x}{2}<\left\lfloor\frac{x}{2}\right\rfloor +1$$
Double and you get:
$$2\left\lfloor\frac{x}{2}\right\rfloor\leq x<2\left\lfloor\frac{x}{2}\right\rfloor +2$$
Taking the floor gives:
$$2\left\lfloor\frac{x}{2}\right\rfloor\leq\lfloor x\rfloor<2\left\lfloor\frac{x}{2}\right\rfloor +2$$
Subtracting gets:
$$0\leq \left\lfloor x\right\rfloor-2\left\lfloor\frac{x}{2}\right\rfloor<2.$$
Which is the result you want.
Part i
You can actually prove the same way  that $\lfloor x+y\rfloor-\lfloor x\rfloor -\lfloor y\rfloor$ is always either $0$ or $1.$ More specifically, you can show that $\lfloor x+y\rfloor-\lfloor x\rfloor -\lfloor y\rfloor=1$ if and only if $\{x\}+\{y\}\geq 1.$ 
In particular, when $y=x,$ $\lfloor 2x\rfloor -2\lfloor x\rfloor=1$ if and only if $\{x\}\geq \frac{1}{2}.$
Now, if $\{x\}+\{y\}\geq 1$ then one or both of $\{x\}$ and $\{y\}$ are $\geq \frac{1}{2}$, so one or both of $\lfloor 2x\rfloor -2\lfloor x\rfloor$ and $\lfloor 2y\rfloor-2\lfloor y\rfloor$ are one. 
So that means, doe any $x,y:$
$$\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor\leq \left(\lfloor 2x\rfloor -2\lfloor x\rfloor\right)+\left(\lfloor 2y\rfloor -2\lfloor y\rfloor\right)$$
This is because when the left side is $0,$ we know the right side is at least $0,$ and when the left side is $1$, then the right side is either $1$ or $2.$
Adding $2\lfloor x\rfloor + 2\lfloor y\rfloor$ to both sides gives you:
$$\lfloor x+y\rfloor +\lfloor x\rfloor +\lfloor y\rfloor\leq \lfloor 2x\rfloor +\lfloor 2y\rfloor$$
A: 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*}
A: 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)
$$
\eqalign{
  & \left\lfloor x \right\rfloor  = \left\lfloor {x/2 + x/2} \right\rfloor  =   \cr 
  &  = 2\left\lfloor {x/2} \right\rfloor  + \left[ {1/2 \le \left\{ {x/2} \right\}} \right] \cr} 
$$
