Smallest possible value of expression involving greatest integer function

Question

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.

Answer 1

Since for all reals $x$ and $y$ we have $$[x]+[y]+1\geq[x+y],$$ by AM-GM we obtain $$\sum_{cyc}\left[\frac{a+b}{c}\right]\geq\left[\sum_{cyc}\frac{a+b}{c}\right]-2\geq6-2=4.$$

Answer 2

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.