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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.

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  • $\begingroup$ Actually. What is "the largest integer" function. Do you mean largest integer equal or less than? What I grew up calling the "floor function". If so, see my answer. If not, ... forgive my answer. $\endgroup$ – fleablood Dec 1 '17 at 7:28
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Since for all reals $x$ and $y$ we have $$[x]+[y]+1\geq[x+y],$$ by AM-GM we obtain $$\sum_{cyc}\left[\frac{a+b}{c}\right]\geq\left[\sum_{cyc}\frac{a+b}{c}\right]-2\geq6-2=4.$$

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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.

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  • $\begingroup$ But the answer provided to me is 4 $\endgroup$ – Rohan Shinde Dec 1 '17 at 6:58
  • $\begingroup$ Really? Not sure how I could be wrong. Are you sure it is the ceiling function? $\endgroup$ – fleablood Dec 1 '17 at 6:59
  • $\begingroup$ What is a ceiling function and by the way I just rechecked the question and the question I posted seems to be correct according to the textbook $\endgroup$ – Rohan Shinde Dec 1 '17 at 7:00
  • $\begingroup$ greatest integer function I thought ... oh, you mean floor function. My mistake. If n < x < n+1 the floor function is n. the ceiling is n+1. I assumed you meant ceiling. $\endgroup$ – fleablood Dec 1 '17 at 7:02
  • $\begingroup$ It was my mistake. I read "greatest integer function" and took it to mean "the least integer that is greater than or equal to". Obviously, you meant "the greatest integer that is less than or equal to". $\endgroup$ – fleablood Dec 1 '17 at 7:33

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