Given $x_1,x_2,...,x_n$ real numbers, prove that there exists a real number $y$, such that, $$\{y-x_1\}+\{y-x_2\}+...+\{y-x_n\} \leq \frac{n-1}{2}$$
Problem
Source: JBMO TST 3 2017 P2
Tags: inequalities
TuZo
14.11.2018 22:57
AlastorMoody wrote: Given $x_1,x_2,...,x_n$ real numbers, prove that there exists a real number $y$, such that, $$(y-x_1)+(y-x_2)+...+(y-x_n) \leq \frac{n-1}{2}$$ Nonsens, for example $y=\sum\limits_{i=1}^{n}{{{x}_{i}}}+\frac{n-1}{2}-k$ with $k$ positive or zero.
XbenX
18.06.2019 16:11
Lemma: For $a,b$ real numbers holds $\{a-b\}+\{b-a\}\in \{0,1\}$, therefore $\{a-b\}+\{b-a\} \leq 1$
Proof: Trivial.
Assume that for $y=x_i, 1\leq i \leq n-1$ we have $\sum_{i=1}^{n} \{y-x_i \}>\frac{n-1}{2}$
Adding these sums we have that $\sum_{1\leq i<j \leq n-1} \left(\{x_i-x_j\}+\{x_j-x_i\}\right)+\sum_{i=1}^{n-1} \{x_i-x_n \}>\frac{(n-1)^2}{2}$
Using the lemma we have
$\frac{(n-1)^2}{2}\leq n-1-\sum_{i=1}^{n-1} \{x_n -x_i \}+\dbinom{n-1}{2} \Longleftrightarrow \sum_{i=1}^{n-1} \{x_n -x_i \}\leq \frac{n-1}{2}.$ $\blacksquare$
electrovector
11.04.2020 15:25
I am not sure if this is correct but it is very similar to @XbenX's solution.
Assume that $\forall i = 1, 2, …, n$ $\sum_{j=1}^n \{x_i-x_j\} > \frac{n-1}{2}$.
Adding these sums we have that $\sum_{1\leq i<j \leq n} \left(\{x_i-x_j\}+\{x_j-x_i\}\right) > \frac{n(n-1)}{2}$ .
Given lemma says $ \frac{n(n-1)}{2}=\dbinom{n}{2} \ge LHS > \frac{n(n-1)}{2}$. Contradiction.