A math inequality cannot prove I try to solve a inequality, I use R to run simulation with different $\alpha>0$ and a sequence of $n$, the simulation all show the LHS is smaller than RHS, but I cannot prove it analytically, can anyone help? Thx!
$$\sum_{j=1}^n \frac{n}{(j^{2\alpha+1}+n)^2}\leq n^{\frac{-2\alpha}{2\alpha+1}}\quad \alpha>0$$
I attach the code:

simulation<-function(alpha){
    result=NULL
    for(n in 1:100){
      j<-1:n
      denom=(j^(2*alpha+1)+n)^2
      result=c(result,sum(n/denom)-n^(-2*alpha/(2*alpha+1)))
    }
    sum(result>0)
  }
  sapply(seq(0.1,10,by=0.1), simulation)  # all zero
  

 A: Not a full solution.
Here is an idea which you might want to exploit.
Consider the function $f(x) = \frac{n}{(x + n)^2}$. We see that $f(x)$ is monotonously decreasing, convex and approaching $0$ as $x \to \infty$, and considerably "smooth".
Now interprete $\frac{n}{(j^{2\alpha+1}+n)^2}$ as probes of $f(x)$, albeit at non-equidistant points $x \in \{1^{2\alpha+1}, 2^{2\alpha+1}, \cdots , n^{2\alpha+1}\}$. 
The crucial part now is to establish that you may estimate an upper bound to the sum of these probes by a sum of more probes which are  equidistant, and proper scaling. I.e. take probes at points $x \in \{1,2,3, \cdots , n^{2\alpha+1}\}$ where now you have taken $n^{2\alpha+1}$ probes (modulo 1) instead of $n$ many, so the scaling factor is $\frac{n}{n^{2\alpha+1}}$. So the idea is to establish (if possible / or in modified form)  first:
$$\sum_{j=1}^n \frac{n}{(j^{2\alpha+1}+n)^2}\leq \frac{n}{n^{2\alpha+1}} \sum_{j=1}^{n^{2\alpha+1}} \frac{n}{(j+n)^2}$$
Once this is done, you can continue with integral bounds:
$$\sum_{j=1}^n \frac{n}{(j^{2\alpha+1}+n)^2}\leq \frac{n}{n^{2\alpha+1}} \sum_{j=1}^{n^{2\alpha+1}} \frac{n}{(j+n)^2}\\
\leq \frac{n}{n^{2\alpha+1}} \int_{x=1}^{n^{2\alpha+1}} \frac{n}{(x+n)^2} {\rm{d}} x \\
=\frac{n^2}{n^{2\alpha+1}}\left[ \frac{1}{n+1} - \frac{1}{n+n^{2\alpha+1}}\right] \color{red}{\le} n^{\frac{-2\alpha}{2\alpha+1}}
 $$
and the last (${\color{red}{\rm{red}}}$) inequality seems to hold true for all $n$ and $\alpha$ as far as I can see (from simulations).
