least value of $\lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{c+b}{a}\rfloor+\lfloor \frac{a+c}{b}\rfloor$ 
If $a,b ,c>0$ . Then least value of $$\bigg\lfloor \frac{a+b}{c}\bigg\rfloor+\bigg\lfloor \frac{c+b}{a}\bigg\rfloor+\bigg\lfloor \frac{a+c}{b}\bigg\rfloor$$

Where $\lfloor x\rfloor$ is floor function of $x$
Plan
Using $$x-1< \lfloor x\rfloor\leq x$$
$$\frac{a+b}{c}-1< \bigg\lfloor \frac{a+b}{c}\bigg\rfloor \leq   \frac{a+b}{c}$$
$$\frac{b+c}{a}-1< \bigg\lfloor \frac{b+c}{a}\bigg\rfloor \leq\frac{b+c}{a}$$
$$\frac{c+a}{b}-1< \bigg\lfloor\frac{c+a}{b}\bigg\rfloor \leq \frac{c+a}{b}$$
How do i solve it Help me please

$$= 3 \bigg\lfloor \frac{a+b}{c} \bigg\rfloor$$
And given all variables are positive, the minimum occurs when $a+b$ is as close to $0$ as possible and $c$ is as large as possible.
