least value of $\lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{c+b}{a}\rfloor+\lfloor \frac{a+c}{b}\rfloor$ If $a,b ,c&gt;0$ . 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 $$x-1&lt; \lfloor x\rfloor\leq x$$

$$\frac{a+b}{c}-1&lt; \bigg\lfloor \frac{a+b}{c}\bigg\rfloor \leq   \frac{a+b}{c}$$

$$\frac{b+c}{a}-1&lt; \bigg\lfloor \frac{b+c}{a}\bigg\rfloor \leq\frac{b+c}{a}$$

$$\frac{c+a}{b}-1&lt; \bigg\lfloor\frac{c+a}{b}\bigg\rfloor \leq \frac{c+a}{b}$$

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.
$$= 3 \bigg\lfloor \frac{a+b}{c} \bigg\rfloor$$

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.