floor sum
16 dez 2011, 15:25
If \(a,b,c>0\) then smallest possible value of
\(\displaystyle \lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{b+c}{a}\rfloor+\lfloor \frac{c+a}{b}\rfloor\)
\(\lfloor .\rfloor =\) floor function
\(\displaystyle \lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{b+c}{a}\rfloor+\lfloor \frac{c+a}{b}\rfloor\)
\(\lfloor .\rfloor =\) floor function
Re: floor sum
16 dez 2011, 16:15
I suppose it's 6... but I'm not sure...
Re: floor sum
16 dez 2011, 16:17
No, it might be 5 
When a=0,5 b=0,5 and c=0,9 for instance you get
1+2+2=5
Less than 5 I can't find the numbers but it might be possible...
When a=0,5 b=0,5 and c=0,9 for instance you get
1+2+2=5
Less than 5 I can't find the numbers but it might be possible...
Re: floor sum
18 dez 2011, 17:06
It is 4. That can be achieved with a=2, b=2, c=2-\(\epsilon\)
It is easy to see that if the 3 numbers are not close enough to eachother, at least one of the terms of the
will be too high.
But if they are all equal, we are in the condition that the floor is actually the number we are considering.
We should think of ways to get numbers where the floor will take most of the terms almost one unit down.
By using small variations on a, b and c, we see that we can obtain 4.
Could we obtain 3? In that case we would get, at best, 1 in each floor calculation. Is that possible? No. Just try! It is easy to show!
It is easy to see that if the 3 numbers are not close enough to eachother, at least one of the terms of the
will be too high.But if they are all equal, we are in the condition that the floor is actually the number we are considering.
We should think of ways to get numbers where the floor will take most of the terms almost one unit down.
By using small variations on a, b and c, we see that we can obtain 4.
Could we obtain 3? In that case we would get, at best, 1 in each floor calculation. Is that possible? No. Just try! It is easy to show!