Showing that $\sum_{i=1}^n \frac{1}{|x-p_i|} \leq 8n \left( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} \right)$ I'm taking a summer analysis course and preparing for our final exam later this week. Our professor gave us the following problem on our mock exam, and I can't seem to get anywhere on it. Does anyone have an idea of how one might proceed? I've tried thinking about it in terms of a geometric series, but that got me nowhere. I've also studied the behavior of the first several partial sums, but again to no avail.
Let $0 \leq p_i \leq 1$ for $i = 1, 2, \dots, n$. Show that
$$ \sum_{i=1}^n \frac{1}{|x-p_i|} \leq 8n \left( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} \right)$$
for some $x$ satisfying $0 \leq x \leq 1$.

I'll do a sloppy version, where I don't pay much attention to constants and don't answer the question fully. A more careful version might lead to a complete answer though.
Let $\delta=1/(4n)$. Then
$$
A=\{ x\in [0,1]: |x-p_j|\ge\delta \:\textrm{ for all }j=1,\ldots, n \}
$$
has measure $|A|\ge 1/2$. Also, notice that
$$
\int_{|x-p|\ge\delta} \frac{dx}{|x-p|} < -2\ln\delta .
$$
Thus $\int_A \sum |x-p_j|^{-1}\, dx < 2n\ln 4n$. It follows that there is an $x\in A$ that makes the sum $<4n\ln 4n = 4n\ln n +O(n)$, which is also the asymptotic size of your bound.
In fact, we can getter a better asymptotic bound if we take $\delta=\epsilon/n$ with $\epsilon>0$ small. This leads to a bound of the form $(2+\epsilon')n\ln n + O(n)$.
