Find the smallest possible value of the expression \[\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor\] in which $a,~ b,~ c$, and $d$ vary over the set of positive integers. (Here $\lfloor x\rfloor$ denotes the biggest integer which is smaller than or equal to $x$.)
Problem
Source: Benelux MO 2014 Problem 1
Tags: function, floor function, algebra unsolved, algebra
17.07.2014 21:38
Denote the expression by $f(a,b,c,d)$. Let $g(a,b,c,d)=\frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}$. Then, by Cauchy-Schwarz, $g(a,b,c,d)=(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})-4 \ge 16-4=12$. Now, we see that $f(a,b,c,d) \le g(a,b,c,d)$, but the difference must be less than 4. But because $f(a,b,c,d)$ is an integer, we know $f(a,b,c,d) \ge 9$. Now choose $a=10$, $b=c=d=11$ to see that 9 is indeed the minimum.
02.05.2016 01:53
If $a_1,a_2,\cdots,a_n \in N^*$$(n\ge 3)$ and $S=a_1+a_2+\cdots+a_n$ . Then$$ \left\lfloor\frac{S-a_1}{a_1}\right\rfloor+\left\lfloor\frac{S-a_2}{a_2}\right\rfloor +\cdots+\left\lfloor\frac{S-a_n}{a_n}\right\rfloor\ge (n-1)^2.$$n=3
21.01.2022 20:12
Tintarn wrote: Denote the expression by $f(a,b,c,d)$. Let $g(a,b,c,d)=\frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}$. Then, by Cauchy-Schwarz, $g(a,b,c,d)=(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})-4 \ge 16-4=12$. Now, we see that $f(a,b,c,d) \le g(a,b,c,d)$, but the difference must be less than 4. But because $f(a,b,c,d)$ is an integer, we know $f(a,b,c,d) \ge 9$. Now choose $a=10$, $b=c=d=11$ to see that 9 is indeed the minimum. Nice solution!
22.01.2022 05:08
sqing wrote: If $a_1,a_2,\cdots,a_n \in \mathbb{N}^*$$(n\ge 3)$ and $S=a_1+a_2+\cdots+a_n$ . Then$$ \left\lfloor\frac{S-a_1}{a_1}\right\rfloor+\left\lfloor\frac{S-a_2}{a_2}\right\rfloor +\cdots+\left\lfloor\frac{S-a_n}{a_n}\right\rfloor\ge (n-1)^2.$$n=3 This problem can be also solved by this method.
30.03.2022 17:51
By a property of floor functions and AM-GM: $$\sum_{\text{cyc}}\left\lfloor\frac{a+b+c}d\right\rfloor\ge\sum_{\text{cyc}}\left(\frac{a+b+c}d-1\right)=\sum_{\text{cyc}}\frac{a+b+c}d-4=\sum_{\text{sym}}\frac ab-4\ge8.$$Equality should be when $a=b=c=d$, but this gives a value of $12$. So this inequality is actually strict, and in fact $\sum_{\text{cyc}}\left\lfloor\frac{a+b+c}d\right\rfloor\ge9$. Equality is at least when $a=b=c=11$ and $d=9$.