Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$. Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
How should I solve this?  I can't think of a way with casework and I can't really simplify it more.  Thanks in advance for posting a proof!

In another way, premised that:
$$\begin{gathered}
  \left\lfloor {x + y} \right\rfloor  = \left\lfloor {\left\lfloor x \right\rfloor  + \left\{ x \right\} + \left\lfloor y \right\rfloor  + \left\{ y \right\}} \right\rfloor  =  \hfill \\
   = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor  =  \hfill \\
   = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \\ 
\end{gathered}$$
where the square brackets indicate the Iverson bracket 
($\left[ {FALSE} \right] = 0,\;\left[ {TRUE} \right] = 1$)
then
$$\begin{gathered}
  \left\lfloor {2x} \right\rfloor  + \left\lfloor {2y} \right\rfloor  = 2\left\lfloor x \right\rfloor  + 2\left\lfloor y \right\rfloor  + \left[ {1 \leqslant 2\left\{ x \right\} + 2\left\{ y \right\}} \right] \hfill \\
  \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left\lfloor {x + y} \right\rfloor  = 2\left\lfloor x \right\rfloor  + 2\left\lfloor y \right\rfloor  + \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \\ 
\end{gathered}$$
and, clearly
$$\left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \leqslant \left[ {1/2 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right]$$
