Number of local maxima of a function  Let $z_j$ ($j=1,\dots, k$) be $k$ points on the complex plane none of which lies on the real line. Is it always true that the function 
$$ F(x)=\sum_{j=1}^k \frac{1}{|x-z_j|^2} $$ has at most $k$ local maxima on the real line?
 A: $F(x)=c$ has for any value of $c>0$ at most $2k$ solutions, which in a weak way supports the conjecture that the statement is always true.
A: If $z_{j}=a_{j}+i\cdot b_{j}$ then $\left | x-z_{j} \right |^{2}=(x-a_{j})^{2}+b_{j}^{2}$ and $$F(x)=\sum_{j=1}^{k}\frac{1}{(x-a_{j})^{2}+b_{j}^{2}}$$ 
Function $F(x)$ is continuous and differentiable.
$${F}'(x)=\sum_{j=1}^{k}\frac{-2(x-a_{j})}{[(x-a_{j})^{2}+b_{j}^{2}]^{2}}=-2\frac{P(x)}{\prod_{l=1}^{k} [(x-a_{l})^{2}+b_{l}^{2}]^{2}}$$
Where $$P(x)=\sum_{j=1}^{k}(x-a_{j})\prod_{l=1, l\neq j}^{k}[(x-a_{l})^{2}+b_{l}^{2}]^{2}$$
and $grad P(x)=4k-3$ which means there could be maximum $4k-3$ zeros of $P(x)$ or $F(x)$ has $4k-3$ extrema.
On top of this, if we assume the following ordering 
$$a_{1} \leqslant a_{2} \leqslant ... \leqslant a_{k}$$
$$x-a_{1} \geqslant x-a_{2} \geqslant  ... \geqslant  x-a_{k}$$
and $x = a_{1}$
$$0 \geqslant a_{1}-a_{2} \geqslant  ... \geqslant  a_{1}-a_{k}$$
if $x = a_{k}$
$$a_{k}-a_{1} \geqslant a_{k}-a_{2} \geqslant  ... \geqslant a_{k}-a_{k-1} \geqslant 0$$
or basically ${F}'(a_{1}) \geqslant 0$ and ${F}'(a_{k}) \leqslant 0$ (*)
Further, ${F}'(x)$ is positive on $(-\infty ,a_{1}]$, looking at $P(x)$ and considering $x \leqslant a_{1} \leqslant a_{2} \leqslant ... \leqslant a_{k}$, thus ${F}(x)$ is ascending on this interval. ${F}'(x)$ is negative on $[a_{k},\infty )$, thus ${F}(x)$ is descending on this interval.
As a result, all the "fun" happens inside $(a_{1},a_{k})$ and according to (*) there is at least one zero for ${F}'(x)$ on this interval (or at least one extrema for $F(x)$).
The remaining part is to sort out $grad P(x)=4k-3$. Assuming the "oscillating" nature (max, min, max, min ...) within $(a_{1},a_{k})$ for $F(x)$ and ascending/descending nature outside that interval, the number of maxima should be greater than the number of minima by 1. Or $n+1$ - maxima, $n$ - minima and $2n + 1 = 4k - 3$ or $2k - 1$ maxima so far ...
