Show that $[2x]+[2y] \geq [x]+[y]+[x+y]$

Solution 1:

Alternative route:

Write $x=n+r$ and $y=m+s$ where $n$ and $m$ are integers and $r,s\in\left[0,1\right)$.

Then $\lfloor2x\rfloor+\lfloor2s\rfloor=2n+2m+\lfloor2r\rfloor+\lfloor2s\rfloor$ and $\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor=2n+2m+\lfloor r+s\rfloor$.

This shows that it is enough to prove $\lfloor2r\rfloor+\lfloor2s\rfloor\geq\lfloor r+s\rfloor$

For this discern the cases $r,s\in\left[0,0.5\right)$ and $r\in\left[0.5,1\right)\vee s\in\left[0.5,1\right)$.

Solution 2:

Your proof is flawed for the reason described by Adam Hughes in his comment. It is not the case that $$\lfloor x+y \rfloor + 1 = \lfloor x + 1/2 \rfloor + \lfloor y + 1/2 \rfloor$$ from the given inequality, because simply taking the floor of the inequality $$\lfloor x + 1/2 \rfloor + \lfloor y + 1/2 \rfloor \le x + y + 1$$ does not automatically turn the inequality into an equality. A simple counterexample is $x = y = 1/4$. Another counterexample is $x = 3/4$, $y = 1/4$.