TITLE: Show that there exist $k\in\{1,2,\cdots,n\}$ such that $\frac{1}{n}\sum_{i=1}^{n}\left(\{kx_{i}\}-\frac{1}{2}\right)^2>\frac{1}{12}-\frac{1}{6n}$
QUESTION [12 upvotes]: The following question has post mathsatck :Nice problem, perhaps from a question of expectation.The problem seemed so elementary, but I had tried to solve it for a long time and eventually failed.I 'd like to get your help or help out here.Thanks!

Show  that: for any real numbers $x_{1},x_{2},\cdots,x_{n}$, there exist $k\in\{1,2,\cdots,n\}$ such that
  $$\dfrac{1}{n}\sum_{i=1}^{n}\left(\{kx_{i}\}-\dfrac{1}{2}\right)^2>\dfrac{1}{12}-\dfrac{1}{6n},$$
  where $\{x\}=x-\lfloor x\rfloor$.

REPLY [14 votes]: Here's an elementary proof of the inequality
$$
(1) \qquad\qquad
\sum_{k=1}^{N-1} \left(1-\frac{k}{N}\right)B_2(\{kx\})
 \ge
 \frac1{12N} - \frac1{12}.
\qquad\qquad\phantom{(1)}
$$
This is nearly the same as the inequality cited by Joe Silverman
(with $N=K+1$), but with the lower bound improved from $-\frac1{12}$ to
$\frac1{12N} - \frac1{12}$.  This is best possible; equality holds iff
$x = m/N$ for some integer $m$ with $\gcd(m,N)=1$.  The slight
improvement by $\frac1{12N}$ seems to be exactly what's needed to
prove the inequality posed by function sug.
Proof: We prove for any $x_1,\ldots,x_N \in \bf R$ the inequality
$$
(2) \qquad\qquad\qquad
\sum_{i=1}^N \sum_{j=1}^N B_2(\{x_i-x_j\}) \geq \frac1{6N},
\qquad\qquad\qquad\phantom{(2)}
$$
with equality iff the $x_i \bmod 1$ are a permutation of
$\{x_0 + (m/N): 0 \leq m < N\}$ for some $x_0$.
To recover (1) from (2), set $x_k = kx$, remove the terms with $i=j$
(which contribute $N/6$ to the total), and divide by 2N.
Permuting the $x_i$ does not change the double sum in (2).
If the $x_i \bmod 1$ are a permutation of
$\{x_0 + (m/N): 0 \leq m < N\}$ then the sum in (2) is
$N \sum_{i=0}^{N-1} B_2(i/N)$, which indeed equals $1/6N$.
We next show that this is minimal.  Permute the $x_i$ so that
$x_1 \le x_2 \le x_3 \le \cdots \le x_N \le x_1 + 1$, and write
the sum in (2) as
$$
\sum_{i=1}^N \sum_{j=1}^N B_2(\{x_{i+j}-x_j\}),
$$
with the index $i+j$ taken $\!\!\mod N$.
Then for each $i$ the fractional parts $\{x_{i+j}-x_j\}$ in the inner sum
total $i$.  Since $B_2$ is convex upwards, this inner sum is minimized
when $\{x_{i+j}-x_j\} = i/N$ for each $j$.  This is the case when
each $x_m \equiv x_0 + m/N$, and that sufficient condition is
also necessary at least when $i=1$.  This completes the proof.  $\Box$