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.

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.

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.