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!

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).
