Prove that $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$ Prove that $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor \,\,\forall x,y \ge 0$
My attempt so far:
Suppose that either $x$ or $y$ is greater than $0$.
Now $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor 4x\rfloor + \lfloor x\rfloor +\lfloor 4y\rfloor + \lfloor y\rfloor \ge \lfloor 4x\rfloor + \lfloor 4y\rfloor +1\ge \lfloor 4x+4y\rfloor \ge \lfloor 3x+y\rfloor + \lfloor x+3y\rfloor$
For $x,y \in [0,1)$!
Proof that
$\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$
is FALSE.
Let $x=^-1$ and $y=0$ then the inequality above becomes $^-5 + 0 \ge ^-3 + ^-1$; which is FALSE.
Proof that
$\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$
is FALSE.
Let $x=^-1$ and $y=0$ then the inequality above becomes $^-5 + 0 \ge ^-3 + ^-1$; which is FALSE.