How to find the minimum value of $\sum_{1\le i
let $a_{1},a_{2},\cdots,a_{6}$ be real numbers,and such
  $$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=2014$$
Find the minimum of the value
$$\sum_{1\le i<j\le 6}[a_{i}+a_{j}]$$
where $[x]$ is the largest integer not greater than $x$.
My idea: since $$[x]>x-1$$
but this inequality can't solve this problem. this is  Beijing university mathematics in 2014
Thank you
 A: If $x+y+z=2014$, then
$$[x]+[y]+[z]\geq 2012$$
In fact 
$$[x]+[y]+[z]= [x]+[y]+[2014-x-y]=2014+[-\{x\}-\{y\}]\geq 2012$$
so $$[a_1+a_2]+[a_3+a_4]+[a_5+a_6]\geq 2012$$
we have 
$$\sum_{1\le i<j\le 6}[a_{i}+a_{j}]\geq2012\times5=10060$$
Let $a_1=2012+\dfrac13,a_2=a_3=\dotsb=a_6=\dfrac13$, then $\sum[a_{i}+a_{j}]=10060$. 
A: The answer by Clin is correct: the minimum is 10060.
One has only to fill in some details of the derivation of the lower bound
$$
\sum_{1\leq i<j\leq 6}[a_i+a_j] \:\geq\: 5\cdot 2012 \:=\: 10060~.
$$
The missing details are embodied in the identity
$$
\begin{aligned}
\sum\nolimits_{1\leq i<j\leq 6}[a_i+a_j] \:=\:
    &~[a_1+a_2]+[a_3+a_4]+[a_5+a_6]+\\[-.9ex]
    &~[a_1+a_3]+[a_2+a_5]+[a_4+a_6]+\\
    &~[a_1+a_4]+[a_2+a_6]+[a_3+a_5]+\\
    &~[a_1+a_5]+[a_2+a_4]+[a_3+a_6]+\\
    &~[a_1+a_6]+[a_2+a_3]+[a_4+a_5]~.
\end{aligned}
$$
Now one uses the lower estimate, proved by Clin,
$$
[a_{\pi1}+a_{\pi2}]+[a_{\pi3}+a_{\pi4}]+[a_{\pi5}+a_{\pi6}]\:\geq\: 1012
$$
for the five permutations $\pi$
that can be read off the 'rows' on the right hand side of the identity.
Re the tentative solution offered by john mangual.
The tight upper bound for the sum $\sum_{1\leq i<j\leq 6}\{a_i+a_j\}$ is $10$,
so it is indeed smaller than $\binom{6}{2}=15$.
Mark that the fractional parts $\{a_i\}$ are not independent,
they must satisfy the condition that $\sum_{i=1}^6\{a_i\}$ is an integer
$\bigl($and so is then the sum  $\sum_{1\leq i<j\leq 6}\{a_i+a_j\}$$\bigr)$.
A: Using the identity $ [x] + \{ x\} = x$, we have
$$ \sum_{1\le i<j\le 6}[a_{i}+a_{j}] + \boxed{ \displaystyle\sum_{1\le i<j\le 6} \{a_{i}+a_{j}\} }= 
\sum_{1\le i<j\le 6} a_{i}+a_{j} = 5 \cdot 2014
 $$
In order to maximize the middle term it is certainly at most 
$$ \sum_{1\le i<j\le 6} 1  = \tfrac{6\cdot 5}{2} = 15$$ 
How is that,  $5 \cdot 2014 - 15$?
