Showing Inequality using Gauss Function 
If $\alpha, \beta\in \Bbb{R}$ and $m, n\in \Bbb{N}$ show that the inequality
$[(m+n)\alpha]+[(m+n)\beta] \ge [m\alpha]+[m\beta]+[n\alpha+n\beta]$
holds iff m=n

I thought that we have to find a counter-example such that when $m\neq n$, there always exists certain (m, n) which makes the inequality doesn't hold. But it was too confusing to do so. I also tried basic inequalities with Gauss functions, but I was rather afraid I might not be able to find counter-examples in that case.
Could it be possible to break it through using rather simple methods rather than seperating each possible case and find a counter-example in each case?
Thanks!
 A: You can proceed as follows.
Lemma 1. The inequality is true if $m=n$.
Proof of lemma 1. Replacing $m$ with $n$, the inequality becomes
$$
[2n\alpha]+[2n\beta] \geq [n\alpha]+[n\beta]+[n(\alpha+\beta)] \tag{1}
$$
Let us put $x=n\alpha-[n\alpha],y=n\beta-[n\beta]$. Then $x,y$ are in $[0,1)$, $[2n\alpha]=2[n\alpha]+[2x], [2n\beta]=2[n\beta]+[2y]$ and $[n(\alpha+\beta)]=[n\alpha]+[n\beta]+[x+y]$, so that
(1) reduces to  $[2x]+[2y] \geq [x+y]$. Now $[x+y]$ can only be $0$ or $1$ ;
if it is $0$ the inequality is clear, and if it is $1$ then at least one of
$x,y$ is larger than $\frac{1}{2}$, which concludes the proof of Lemma 1.
Lemma 2. If $m < n$, the inequality becomes false  for 
$\alpha=\frac{2m^2+1}{2mn(m+n)}, \beta=\frac{2mn-1}{2mn(m+n)}$.
Proof of lemma 2. Note that $\alpha+\beta=\frac{1}{n}$, 
$(m+n)\alpha=1-\frac{2m(n-m)-1}{2mn}$ and
$(m+n)\beta=1-\frac{1}{2mn}$. We deduce that
$$
[(m+n)\alpha]=[(m+n)\beta]=[n\alpha]=[n\beta]=0, [n(\alpha+\beta)]=1.
$$
Lemma 3. If $m > n$ and $m$ is not a multiple of $n$, the inequality becomes false  for $\alpha=\frac{1}{m}, \beta=\frac{q}{m}$, where $q=[\frac{m}{n}]$.
Proof of lemma 3. Indeed, we have $(m+n)\alpha=1+\frac{n}{m}$, so $[(m+n)\alpha]=1$. Since $m$ is not a multiple of $n$, we have $q <\frac{m}{n}$, so $(m+n)\beta=q+q\frac{n}{m}$ yields $[(m+n)\beta]=q$. Trivially $[m\alpha]=1$
and $[m\beta]=q$ ; also $n(\alpha+\beta)=\frac{n}{m}(q+1)$ and 
$\frac{m}{n} \leq  q+1 \leq \frac{m}{n}+1 < \frac{2m}{n}$, so $[n(\alpha+\beta)]=1$. Finally
$$
[(m+n)\alpha]+[(m+n)\beta]=1+q < 1+q+1= [m\alpha]+[m\beta]+[n\alpha+n\beta]
$$
Lemma 4. If $m > n$ and $m$ is a multiple of $n$, say
$m=tn$ for some integer $t\geq 2$, the inequality becomes false for $\alpha=\frac{3t+1}{2nt(t+1)}$, $\beta=\frac{2t^2-1}{2nt(t+1)}$.
Proof of lemma 4. The following equalities hold :
$$
\begin{array}{rclcl}
(m+n)\alpha &=& 1+\frac{t+1}{2t} &=& 2-\frac{t-1}{2t} \\
(m+n)\beta &=& t-1+\frac{2t-1}{2t} &=& t-\frac{1}{2t} \\
m\alpha &=& 1+\frac{t-1}{2(t+1)} &=& 2-\frac{t+3}{2t+2} \\
m\beta &=& t-1+\frac{1}{2(t+1)} &=& t-\frac{2t+1}{2t+2} \\
n(\alpha+\beta) &=& 1+\frac{1}{2(t+1) } & & \\ 
\end{array}
$$
so 
$$
[(m+n)\alpha]+[(m+n)\beta]=1+(t-1) < 1+(t-1)+1= [m\alpha]+[m\beta]+[n\alpha+n\beta]
$$
A: Ewan Delanoy thankfully gave me an ingeneous proof about my question, which is

$[(m+n)x]+[(m+n)y] \ge [mx]+[my]+[nx+ny]$

where $m, n\in \Bbb{N}$ and $x, y\in \Bbb{R}$. I got inspired (or got hints) from his proof, and here is what I've done through my style of understadning.
Case m=n His proof is just excellent. 
Case n>m We have to find counter-examples for any given $m, n$ where $m\not=n$.  So we have freedom to choose certain $x, y$ for given $m, n$. First, in this case, I'll put
$$ \begin{array}{rclcl} \alpha=\frac{t}{m+n},&\beta=\frac{k}{m+n} \end{array}$$
where $\alpha$, $\beta$ are the decimal fractions of x and y respectively and $0 \le t$, $k <1$. Then we can represent the inequality as
$$[(m+n)\alpha]+[(m+n)\beta] \ge [m\alpha]+[m\beta]+[n\alpha+n\beta]$$
The precise selection of the value of $\alpha$ and $\beta$ was to make the terms in the left side to zero. Moreover, consider
$$t+k=\frac{m}{n}+1$$
t and k which suffice the given condition always exist since $1 \le \frac{m}{n}+1 <2$. Then 
$$[n\alpha+n\beta]=[n \frac{t+k}{m+n}]=[\frac{m+n}{m+n}]=1>0$$
which is a contradiction. The idea was that since $n>m$, it would be plasible to make all the terms containing $m$ into 0 but not the terms with $n$. But for this part maybe his proof is much more concise.
Case m>n Now this part was indeed very tricky since if we make the term $[n\alpha+n\beta]$ into zero, the inequality obviously holds. My strategy was to make things concise at first and then fill up the missing part one by one.
(1): $2n>m$ First of all, to make things concise, let $\alpha=\beta$. Then the inequality becomes
$$2[(m+n)\alpha] \ge 2[m\alpha]+[2n\alpha]$$
Then let $\alpha=\frac{1}{m}$ since $m\not=1$. Then we get
$$2[\frac{m+n}{m}]=2+[\frac{n}{m}]=2 \ge 2+[\frac{2n}{m}]$$
In order to make a contradiction, $[\frac{2n}{m}] \ge 1$, which is only possible when $2n>m$. To summarize, when $2n>m$, put $\alpha=\beta=\frac{1}{m}$, then contradiction will always arise.
(2): $2n<m$ First of all, since we should not make the term $[n\alpha+n\beta]$ into zero, I'll first fix the term into $n$, i.e. $[\alpha+\beta]=1$. Then consider 
$$\begin{array}{rclcl} \alpha=\frac{2}{m},&\beta=\frac{m-2}{m} \end{array}$$ 
This is possible since $m \ge 3$. Then the right side of the inequality becomes $m+n$. And the left side becomes $m+n-1$, because
$$[\begin{array}{rclcl} \frac{2m+2n}{m}]=2+[\frac{2n}{m}]=2,&
[\frac{(m+n)(m-2)}{m}]=[m+(n-2)-\frac{2n}{m}]=m+n-3 \end{array}$$
thus again contradiction arises. 
(3): $2n=m$ Again, let $\alpha=\beta$. Then the equality becomes
$$2[3n\alpha] \ge 3[2n\alpha]$$
Let $\frac{1}{2n} \le \alpha < \frac{2}{3n}$, then the left side becomes 2, but the right side becomes 3, which is again contradiction.
I tried to give a rather easy way to understand this problem and what kind of strategy I used to solve this. Overall, my strategy was to assume that counter-examples exist in 'dense', so making things more concise would not bother much for me to find certain counter-example for each different cases. Actually, this problem took me almost a day to understand and tackle it, but I think it was a very useful for me to understand and prove inequality based on Gauss function (and finding counter-examples as well).
