the least possible value for :$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $ If we know that for every $a,b,c>0$ ,how we can find  the least possible value for :
$$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $$
 A: The expression
$$
E := \frac {a + b} c + \frac {b + c} a + \frac {c + a} b
$$
is homogeneous, so we can assume $a + b + c = 1$. Then it becomes
$$
E = \frac 1 a + \frac 1 b + \frac 1 c - 3 
$$
By CS inequality we get
$$
\frac 1 a + \frac 1 b + \frac 1 c \geq \frac {(1 + 1 + 1)^2} {a + b + c} = 9
$$
So $E \geq 6$. By the following relation
$$
\left\lfloor \frac {a + b} c \right\rfloor > \frac {a + b} c - 1
$$
(and the other ones got by cyclic permutations of variables), we have
$$
E' := \left\lfloor \frac {a + b} c \right\rfloor + \left\lfloor \frac {b + c} a \right\rfloor + \left\lfloor \frac {c + a} b \right\rfloor > E - 3 \geq 6 - 3 = 3
$$
Being $E'$ an integer, the above inequality is equivalent to
$$
E' \geq 4
$$
To conclude, let's note that for $a = b = 4$ and $c = 3$, $E' = 4$.
A: Put $(a,b,c)=(3,4,4)$ to get 4. I will show this is optimal.
Assume without loss of generality that $a \leq b \leq c$. Two cases:
If $c \geq a+b$ then $\lfloor\frac{c+a}{b}\rfloor \geq 1$ and $\lfloor\frac{c+b}{a} \rfloor \geq \lfloor\frac{a+b+b}{a} \rfloor \geq 3$ so the sum is at least 4.
If $c \leq a+b$ then $\lfloor\frac{a+b}{c}\rfloor \geq 1$, $\lfloor\frac{b+c}{a}\rfloor \geq 2$, $\lfloor\frac{c+a}{b}\rfloor \geq 1$ so the sum is at least 4.
A: I would come up with one line proof starting from the trivial fact that $\left\lfloor x \right\rfloor>x-1$. Then
$$\lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor>\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}-3\ge3.$$
Since the left side is an integer, then the smallest value is $4$.
Q.E.D. 
