Fix a positive integer $n\geq 2$. What is the lest value that the expression $$\bigg\lfloor\frac{x_2+x_3+\dots +x_n}{x_1}\bigg\rfloor + \bigg\lfloor\frac{x_1+x_3+\dots +x_n}{x_2}\bigg\rfloor +\dots +\bigg\lfloor\frac{x_1+x_2+\dots +x_{n-1}}{x_n}\bigg\rfloor$$may achieve, where $x_1,x_2,\dots ,x_n$ are positive real numbers.
Problem
Source: Stars of Mathematics 2024 P1 (junior level)
Tags: inequalities, floor function
SatisfiedMagma
07.12.2024 20:03
Pretty nice problem for the juniors. The answer is $(n-1)^2$. Credit where its due, found the answer and proved it with MrOreoJuice. Basically, homogenize and assume $\sum x_i = 1$ and then Cauchy to find lower bound. Then a construction like below is fine.
Maximilian113
07.12.2024 20:37
Since $\lfloor x \rfloor > x-1$ for all real $x$ we have that $$\bigg\lfloor\frac{x_2+x_3+\dots +x_n}{x_1}\bigg\rfloor + \bigg\lfloor\frac{x_1+x_3+\dots +x_n}{x_2}\bigg\rfloor +\dots +\bigg\lfloor\frac{x_1+x_2+\dots +x_{n-1}}{x_n}\bigg\rfloor > \frac{x_2+x_3+\dots +x_n}{x_1}+ \frac{x_1+x_3+\dots +x_n}{x_2}+\dots +\frac{x_1+x_2+\dots +x_{n-1}}{x_n}-n,$$which by AM-GM on each pair $\frac{x}{y}+\frac{y}{x}$ is greater or equal to $$\frac{n(n-1)}{2} \cdot 2 - n = n^2-2n.$$Now, it suffices to show that $(n-1)^2$ is attainable. To do this, just let the first $n-1$ $x_i$ be $2n-1$ and the remaining one to be $2n-2.$
DensSv
14.12.2024 19:39
Denote $S$ the sum in the statement. Then $$S+n=\bigg\lfloor\frac{x_1+x_2+x_3+\dots +x_n}{x_1}\bigg\rfloor + \bigg\lfloor\frac{x_1+x_2+x_3+\dots +x_n}{x_2}\bigg\rfloor +\dots +\bigg\lfloor\frac{x_1+x_2+\dots +x_{n-1}+x_n}{x_n}\bigg\rfloor$$$$S+n=(x_1+x_2+\cdots+x_n)(\sum_{i=1}^{n}\frac{1}{x_i})-\sum_{i=1}^{n}\biggl\{ \frac{x_1+x_2+x_3+\dots +x_n}{x_i}\biggl\}$$By Bergstrom $$(x_1+x_2+\cdots+x_n)(\sum_{i=1}^{n}\frac{1}{x_i})\geq n^{2}$$Now
$$S\geq n^{2}-n-\sum_{i=1}^{n}\biggl\{ \frac{x_1+x_2+x_3+\dots +x_n}{x_i}\biggl\}$$However, $\sum_{i=1}^{n}\biggl\{ \frac{x_1+x_2+x_3+\dots +x_n}{x_i}\biggl\}<n$. Thus $S>n^{2}-2n$, hence $S\geq (n-1)^{2}$ (since $S\in\mathbb{Z}$).
$S=(n-1)^{2}$ is achived when $x_{1}=n$ and $x_{2}=x_{3}=\cdots=x_{n}=n+1$
I got 5 points for it. During the contest I was certain that the minimum is $n(n-1)$ and I assumed that my solution was missing something, hence I missed the $S=(n-1)^{2}$ construction.
AshAuktober
31.12.2024 10:48
We claim the minimum is $\boxed{(n-1)^2 = n^2-2n+1}$. Firstly note that it is achieved by $x_1 = 1, x_i = 1+e \forall i > 1$, where $e = \frac{1}{1434n} < \frac{1}{n-1}$. This is because any term in the expression with $x_1$ in the numerator will then be $$\left\lfloor\frac{(n-1) + (n-2)e}{1+e} \right \rfloor = n-2,$$and any term with $x_1$ in the denominator will be of the form $$\left\lfloor \frac{(n-1)(1+\epsilon)}{1} \right \rfloor = n-1.$$Therefore, since the expression contains $n-1$ terms of the first type and one term of the second, it evaluates to $$(n-2)(n-1) + n-1 = (n-1)^2,$$as desired. Now we prove the bound. Note that we have $\lfloor x \rfloor > x-1$. Applying this, $$\bigg\lfloor\frac{x_2+x_3+\dots +x_n}{x_1}\bigg\rfloor + \bigg\lfloor\frac{x_1+x_3+\dots +x_n}{x_2}\bigg\rfloor +\dots +\bigg\lfloor\frac{x_1+x_2+\dots +x_{n-1}}{x_n}\bigg\rfloor >\left( \sum_{cyc} \frac{x_2 + x_3 + \dots + x_n}{x_1} \right)- n$$$$ = (x_1 + \dots + x_n)\left(\frac 1{x_1} + \dots + \frac{1}{x_n}\right) - 2n \ge n^2 - 2n, $$where the last part is by Cauchy-Schwarz. Therefore, as it is an integer, $$\bigg\lfloor\frac{x_2+x_3+\dots +x_n}{x_1}\bigg\rfloor + \bigg\lfloor\frac{x_1+x_3+\dots +x_n}{x_2}\bigg\rfloor +\dots +\bigg\lfloor\frac{x_1+x_2+\dots +x_{n-1}}{x_n}\bigg\rfloor \ge n^2 - 2n+1$$as desired. $\square$