TITLE: Show that $\sum_{i=1}^n \frac{1}{|x-p_i|}\le 8n\sum_{i=1}^n \frac{1}{2i-1}$ for some $x,0\le x \le 1$
QUESTION [12 upvotes]: Let $0\le p_i \le 1$ for $i = 1,2,\dots n.$ Show that $\displaystyle\sum_{i=1}^n \dfrac{1}{|x-p_i|}\le 8n\displaystyle\sum_{i=1}^n \dfrac{1}{2i-1}$ for some $x,0\le x \le 1.$ 

First, it seems reasonable to convert the RHS to a closed form. I know $\dfrac{\pi}{4}=\displaystyle\sum_{i=1}^\infty (-1)^{i-1}\dfrac{1}{2i-1},$ but I don't think that will be useful for this problem. Also, I don't think the LHS can be simplified using a telescoping series. I think I should pick several candidates $x_j$ and then take $\displaystyle\sum_{j=1}^n \dfrac{1}{|x_j-p_j|}.$ I should probably also consider different intervals of $[0,1].$
Any help would be appreciated.

REPLY [3 votes]: Let
$$ \Vert x\Vert\, :=\, \min(x\,\, 1-x) $$
so that $\, \forall_{x\in[0;1]}\, \Vert x\Vert\, \le\, x\, $
and
$$ \sum_{i=1}^n \dfrac{1}{\Vert x-p_i\Vert}\, \ge
    \, \sum_{i=1}^n \dfrac{1}{|x-p_i|} $$

The "norm" $\, \Vert\, .\Vert\, $ is more regular (geometric, homogenous) than $\, |\ .|,\, $ and the former of the two leads useful implications about the latter one; consider $\, \mathbb R/\mathbb Z$.

With respect to $\, \Vert\, .\Vert,\, $ the unique (up
to isometry with respect to the distance
$\, d(x\ y):=\Vert x-y\Vert)\, $critical sequence is
$$ \forall_{k=1}^n\quad
     q_k\, :=\, \frac{2\cdot k-1}{2\cdot n} $$
and the critical $\, x\ $ is $\ x:=0\, $ (sure,
any $\, x:=\frac kn\, $ would do too  (for
$\, n=1\ldots n)\, $ where critical means that
$$ \sum_{i=1}^n \dfrac{1}{\Vert x-q_i\Vert}\, \ge
    \, \sum_{i=1}^n \dfrac{1}{\Vert y_m-p_i\Vert} $$
where $\, y_m\, $ provides the minimum of the RHS for the arbitrarily given sequence $\, p_1\ldots p_n\, $ (point $\, y_m\, $ depends on sequence
$\, p_k$).
Numerically, for critical values:
$$ \sum_{i=1}^n \dfrac{1}{\Vert x-q_i\Vert}\,\, =\,
    \, 2\cdot n\cdot\sum_{k=1}^n\frac 1
      {\min(2\cdot k-1,\,\ 2\cdot(n-k)+1)} $$
Obviously,
$$ 2\!\cdot\! n\cdot\sum_{k=1}^n\frac 1
      {\min(2\cdot k-1,\,\ 2\cdot(n-k)+1)}\, \le
   \, 4\!\cdot\!n\cdot \sum_{k=1}^n\frac 1{2\cdot k-1} $$
This beats the required inequality over two times over.