For what $n\in \mathbb{N}$ is $[nx]+[ny] \ge [x]+[y]+[(n-1)(x+y)] \, \forall x,y \in \mathbb{R}$? 
For what $n\in \mathbb{N}$ is $[nx]+[ny] \ge [x]+[y]+[(n-1)(x+y)] \,\, \forall x,y \in \mathbb{R}$ ? Where [x] denotes the greatest integer function/ floor function(denoted by $\lfloor x\rfloor$) of $x$. 

My solution: consider $x=y=\dfrac{1}{n+1}$. For this choice of $x$ and $y$, $[nx]+[ny]=\left[\dfrac{n}{n+1}\right]+\left[\dfrac{n}{n+1}\right]= 0 $ as $n+1\le n \, \forall \,n \in \mathbb{N}$, and $[x]+[y]+[(n-1)(x+y)] =\left[\dfrac{1}{n+1}\right]+\left[\dfrac{1}{n+1}\right]+\left[\dfrac{2(n-1)}{n+1}\right] = 1$ for $2(n-1)\ge (n+1)$ Which gives us $n \ge 3$.  So the statement is not true for $n\ge 3$, and $n=1$ is trivially true. 
Now the only case left is $n=2$. My claim is that the above statement is also true for $n=2$. I was able to prove it by taking $x= I_1+f_1$ and $y=I_2+f_2$, where $I_1,I_2\in \mathbb{Z} $ and $f_1,f_2 \in [0,1)$ denote the fractional part of $x$ and $y$. But it requires taking different cases on $f_1$ and $f_2$, is there a better way to prove for $n=2$ case? 
 A: Your question about "a better way" is not completely clear to me. I'm not quite sure if you mean not requiring taking any cases, a shorter method, and/or something else to prove
$$\lfloor nx\rfloor + \lfloor ny\rfloor \ge \lfloor x\rfloor + \lfloor y\rfloor + \lfloor (n-1)(x+y)\rfloor \,\, \forall \; x,y \in \mathbb{R} \tag{1}\label{eq1A}$$
is true for $n = 2$. Here is what I would do which I believe minimizes the use of cases. Using your idea of $x = I_1 + f_1$ and $y = I_2 + f_2$, where $I_1,I_2\in \mathbb{Z} $ and $f_1,f_2 \in [0,1)$, the LHS of \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned}
\lfloor 2x\rfloor + \lfloor 2y\rfloor & = \lfloor 2I_1 + 2f_1 \rfloor + \lfloor 2I_2 + 2f_2\rfloor \\
& = 2I_1 + 2I_2 + \lfloor 2f_1 \rfloor + \lfloor 2f_2 \rfloor
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The RHS of \eqref{eq2A} becomes
$$\begin{equation}\begin{aligned}
\lfloor x\rfloor + \lfloor y\rfloor + \lfloor (n-1)(x+y)\rfloor & = \lfloor I_1 + f_1 \rfloor + \lfloor I_2 + f_2\rfloor + \lfloor (I_1 + f_1) + (I_2 + f_2) \rfloor \\
& = I_1 + I_2 + \lfloor (I_1 + I_2) + (f_1 + f_2) \rfloor \\
& = 2I_1 + 2I_2 + \lfloor f_1 + f_2 \rfloor
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $0 \le f_1 + f_2 \lt 2$, you get $\lfloor f_1 + f_2 \rfloor$ is either $0$ or $1$. If it's $0$, then as you can see from \eqref{eq2A} as it's $\ge 2I_1 + 2I_2$, \eqref{eq1A} will be true. With it being $1$ instead, this means $f_1 + f_2 \ge 1$, so $f_1 \ge 0.5 \implies 2f_1 \ge 1$ and/or $f_2 \ge 0.5 \implies 2f_2 \ge 1$. From \eqref{eq2A}, this means the LHS is $\ge 2I_1 + 2I_2 + 1$, so \eqref{eq1A} is also true in this case.
I'm not sure if this is much simpler, if at all, than what you did, but I hope it helps you somewhat. I don't think you'll find any method which which you will consider to be a much, if any, "better way" than what you did and what I've done here.
