Show this sum is uniformly bounded in $N$ and in $i$. For $N>0, \ d>0$, I am considering $N$ points $(y_1,...,y_N)$ in $(\mathbb{R}^d)^N$ such as there exists a constant $c>0$ with :
$$\underset{1 \leq i,j \leq N \atop i \neq j}{\min} |y_i-y_j| \geq c, \quad \forall N>0.$$
I am trying to show that there exists a constant $M>0$ such as :
$$\underset{1 \leq i \leq N}{\sup} \sum_{j=1}^N \frac{1}{(c+|y_i-y_j|)^4} \leq M, \quad \forall N>0$$
I have tried to use my hypothesis which gives me :
$$\frac{1}{(c+|y_i-y_j|)^4} \leq \frac{1}{16 c^4}$$
but this leads me nowhere since $(\sum_{j=1}^N \frac{1}{16 C^4})_{N \in \mathbb{N}}$ is divergent.
Update 1
I've proposed a solution to this question, I'm not sure weither it's true or false, if anyone wants to proof read it I would be very thankful.
 A: Okay I think I found a solution to my question. It's actually harder than expected. Anyone who wants to verify it is gladly welcomed.
First of all, notice that the property we want to found is invariant by translation (i.e we can consider $z_i=y_i-Z$ for any $Z \in \mathbb{R}^d$). Therefore we can consider that $i=1$ and $y_1=0$ without loosing generality.
First let's work for a fixed $N>1$. We can also rearranged the indices such as $0=|y_1| < |y_2| < \dots <|y_N|$.
Because of our assumption on the minimal distance, the balls $B(y_i,c/2)$ and $B(y_j,c/2)$ do not intersect for $i \neq j$. Moreover, for any $2<k<N$, we have :
$$\bigcup_{j=1}^k B(y_j,c/2) \subset B(0, c/2 + \sup_{1 \leq j \leq n} |y_j|) \subset B(0,2|y_k|), \quad \text{(a drawing helps)}$$
Now we compute the volumes which gives us :
$$k (\frac{c}{2})^3 \leq (2 |y_k|)^3, \ \ i.e \ \ \frac{c}{4} k^{\frac{1}{3}} \leq |y_k|$$
Last step is using a comparaison between series and integral :
\begin{aligned}
\sum_{j=1}^N \frac{1}{(c+|y_1-y_j|)^4}& \leq \frac{1}{c^4} + \sum_{j=2}^N \frac{1}{|y_1-y_j|^4}  \\
& \leq \frac{1}{c^4} +  \sum_{j=2}^N \frac{1}{|y_j|^4} \quad \text{since $y_1=0$}\\
& \leq \frac{1}{c^4} +\frac{4^4}{c^4} \sum_{i=2}^N i^{-\frac{4}{3}}\\
& \leq \frac{1}{c^4} +\frac{4^4}{c^4} \int_{0}^N x^{-\frac{4}{3}} \ \mathrm{d}x\\
& \leq \frac{1}{c^4} +\frac{\alpha}{N^{\frac{1}{3}}}
\end{aligned}
where $\alpha$ is a positive constant independant of $N$. We finally can use that $\frac{\alpha}{N^{\frac{1}{3}}} \leq \alpha$ since $N>1$, which gives us $M:=\frac{1}{c} + \alpha$ such as
$$\sum_{j=1}^N \frac{1}{(c+|y_1-y_j|)^4} \leq M$$
This computation holds for any $N>1$ and $i \in \{1,\dots,N \}$.
A: I think I found a second answer which is easier. I'm assuming $d=3$ for this answer.
Let fix $i$ between $1$ and $N$. We found easily that there exists a constant $C$ independant of $i$ and $N$ such as :
$$\frac{1}{(c + |y_i-y_j|)^4} \leq C \int_{B(y_j,\frac{c}{2})} \frac{1}{(c + |y_i-y|)^4} \mathrm{d}y, \quad \forall j$$
Therefore when we sum :
$$\sum_{j=1}^N \frac{1}{(c + |y_i-y_j|)^4} \leq \sum_{j=1}^N \int_{B(y_j,\frac{c} {2})}\frac{1}{(c + |y_i-y|)^4} \mathrm{d}y$$
or since all the balls $B(y_j,\frac{c} {2})$ are disjoinct we found that :
$$\sum_{j=1}^N \frac{1}{(c + |y_i-y_j|)^4} \leq \int_{\bigcup_{j=1}^N B(y_j,\frac{c} {2})} \frac{1}{(c + |y_i-y|)^4} \mathrm{d}y \leq \int_{\mathbb{R}^3} \frac{1}{(c + |y_i-y|)^4} \mathrm{d}y  = \int_{\mathbb{R}^3} \frac{1}{(c + |y|)^4} \mathrm{d}y  $$
and this integral is well defined since we are on $\mathbb{R}^3$. We then pose $M:=\int_{\mathbb{R}^3} \frac{1}{(c + |y|)^4} \mathrm{d}y$ and the result should follow.
