Smallest possible value of expression involving greatest integer function If $a, b, c \gt 0$ then what is the smallest possible value of 
$$\left[\frac{a+b}{c}\right]+ \left[\frac{b+c}{a}\right] + \left[\frac{c+a}{b}\right]$$ 
where $[.]$ denotes greatest integer function. 
I tried using the AM GM inequality at first but it was not useful. I also tried adding 1 to each bracket and then subtracting 3 from overall to get the same numerator in each bracket. But this too wasn't useful. I don't have much practice of solving the question involving greatest integer function. Somebody please tell me how to deal with this question.

Wolog $a \le b \le c$
$[\frac {b+c}a] \ge [\frac {a+a}a] =2$
And $[\frac {a+c}b] \ge [\frac {a + b}b] \ge [\frac bb] = 1$.
So you can not get less than $3$
If $\frac {a+b}{c} < 1$ then $c > a+b$ and $\frac {c+a}b > \frac {2a + b}b= \frac {2a}b + 1$.  So If $\frac {c+a}b < 2$ then $\frac {2a}b < 1$ so $b > 2a$.  So $\frac {b+ c}a > \frac {2a + a+b}a > \frac {5a}a = 5$.
In other words if $[\frac {a+b}{c}]=0$ and $[\frac {c+a}b] = 1$ then $[\frac {b+c}b] \ge 5$. So you most certainly can not get $3$ as an answer.
So the answer is $4$ or greater.
Can we get four?
Well, if $a = b = c$ then $\frac {a+b}c= \frac {a+c}b=\frac {b+c}a =2$
Well if we can get $\frac {a+b}c$ and $\frac {a+c}b$ just under $2$ and $\frac {b+c}a$ just over $2$ that should do it.
Let $a= .9; b = 1; c= 1.1$ then
$[\frac {a+b}c]=[\frac {1.9}{1.1}] = 1$ and $[\frac {a+c}b]= [\frac 21]$ 
....oops..... we'll have to shave just a whisker off.
Let $a = .9; b=1; c=1.09$ then $[\frac {a+b}c]=[\frac {1.9}{1.09}]=1$ and
$[\frac {a+c}b]= [\frac {1.99}1]=1$ (barely!)$
And $[\frac {b+c}{a}] = [\frac {2.09}{.9}] =2$.
So $[\frac {a+b}c]+[\frac {a+c}b]+[\frac {b+c}{a}]=4$.
$4$ is the smallest.
