least value of $\lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{c+b}{a}\rfloor+\lfloor \frac{a+c}{b}\rfloor$

Question

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

Is your expression correct? Is it really the sum of three identical terms? — Martin R, Jun 12 '19 at 07:35
Or do you mean this https://math.stackexchange.com/q/200289/42969 ? — Martin R, Jun 12 '19 at 07:36

David G. Stork · Answer 1 · 2019-06-12T07:38:19.013

-2

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

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.

edited Jun 12 '19 at 07:38

answered Jun 12 '19 at 07:36

David G. Stork

29,774

1

From the attempted proof I strongly assume that this is a typo in the question. – Martin R Jun 12 '19 at 07:37
@MartinR: Then downvote the question, not a correct answer! – David G. Stork Jun 17 '19 at 14:14

least value of $\lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{c+b}{a}\rfloor+\lfloor \frac{a+c}{b}\rfloor$

1 Answers1