Show that $\sum_{i=1}^n \frac{1}{|x-p_i|}\le 8n\sum_{i=1}^n \frac{1}{2i-1}$ for some $x,0\le x \le 1$ 
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.
 A: 
I think I should pick several candidates $x_j$

Something along those lines. Take an average. If that average is not larger than the right hand side, it follows that there is an $x$ for which the left hand side is not larger than the right hand side.
Now the question is what average we should consider. Obviously considering
$$\int_0^1 \sum_{j = 1}^n \frac{1}{\lvert x - p_j\rvert}\,dx = +\infty$$
doesn't work here.
Thus let's cut out short intervals around the $p_j$ and look at what that yields. For $0 < \varepsilon < 1$ and $p \in [0,1]$ define $A(p,\varepsilon) = \{x \in [0,1] : \lvert x - p\rvert \geqslant \varepsilon\}$, and then set
$$I(\varepsilon) = \bigcap_{j = 1}^n A(p_j,\varepsilon).$$
For small enough $\varepsilon$ this is a union of finitely many nondegenerate intervals, thus
$$M(\varepsilon) = \int_{I(\varepsilon)}dx > 0,$$
and I claim that for $\varepsilon = \frac{1}{4n}$ the average
$$\frac{1}{M(\varepsilon)} \int_{I(\varepsilon)} \sum_{j = 1}^n \frac{1}{\lvert x - p_j\rvert}\,dx$$
is no larger than $8n \sum_{j = 1}^n \frac{1}{2j-1}$, thus there are $x \in I(\varepsilon)$ satisfying the desired inequality.
First, to obtain $I(\varepsilon)$ we remove $n$ intervals of length at most $2\varepsilon$ (at most, not necessarily exactly, because there may be $p_j \in [0,\varepsilon)$ or in $(1-\varepsilon,1]$). These may overlap, so $M(\varepsilon)$ cannot be exactly determined unless the $p_j$ are explicitly given, but we have the simple bound
$$M(\varepsilon) \geqslant 1 - n\cdot 2\varepsilon,$$
in particular $M(\frac{1}{4n}) \geqslant \frac{1}{2}$.
Further, for $0 < \varepsilon \leqslant \frac{1}{4}$ and $p \in [0,1]$ we have
$$\int_{A(p,\varepsilon)} \frac{dx}{\lvert x-p\rvert} \leqslant 2\log \frac{1}{2\varepsilon}.$$
If $\varepsilon < p < 1 - \varepsilon$, then
\begin{align}
\int_{A(p,\varepsilon)} \frac{dx}{\lvert x - p\rvert}
&= \int_0^{p-\varepsilon} \frac{dx}{p-x} + \int_{p + \varepsilon}^1 \frac{dx}{x-p} \\
&= -\log \varepsilon + \log p + \log (1-p) - \log \varepsilon \\
&= 2\log \frac{1}{\varepsilon} + \log \bigl(p(1-p)\bigr) \\
&\leqslant 2\log \frac{1}{\varepsilon} + \log \frac{1}{4} \\
&= 2\log \frac{1}{2\varepsilon}
\end{align}
since $0 \leqslant p(1-p) \leqslant 1/4$.
If $0 \leqslant p \leqslant \varepsilon$, then
\begin{align}
\int_{A(p,\varepsilon)} \frac{dx}{\lvert x-p\rvert}
&= \int_{p+\varepsilon}^1 \frac{dx}{x-p} \\
&= \log (1-p) - \log \varepsilon \\
&\leqslant \log \frac{1}{\varepsilon} \\
&\leqslant 2\log \frac{1}{2\varepsilon},
\end{align}
where in the last inequality we used $\varepsilon \leqslant 1/4$. For $1 - \varepsilon \leqslant p \leqslant 1$ the claim follows similarly (by symmetry, if you want).
Thus we have, setting $\delta = \frac{1}{4n}$,
\begin{align}
\frac{1}{M(\delta)}\int_{I(\delta)} \sum_{j = 1}^n \frac{1}{\lvert x - p_j\rvert}\,dx
&\leqslant 2\int_{I(\delta)} \sum_{j = 1}^n \frac{1}{\lvert x - p_j\rvert}\,dx \\
&= 2\sum_{j = 1}^n \int_{I(\delta)} \frac{dx}{\lvert x - p_j\rvert} \\
&\leqslant 2\sum_{j = 1}^n \int_{A(p_j,\delta)} \frac{dx}{\lvert x - p_j\rvert} \\
&\leqslant 2\sum_{j = 1}^n 2\log \frac{1}{2\delta} \\
&= 4n\log (2n).
\end{align}
Together with
$$\sum_{j = 1}^n \frac{2}{2j-1} \geqslant \sum_{j = 1}^n \int_{2j-1}^{2j+1} \frac{dx}{x} = \log (2n+1) > \log (2n)$$
the assertion follows.
A: 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.
